# Homomorphism Conjugacy versus Centralizer Actions in the Symmetric Group

Suleika Norrbom

###### Abstract

We explore when generator-conjugate homomorphisms are conjugate and when
element-conjugate homomorphisms are conjugate from abelian or dihedral groups
to the symmetric group. We completely determine when such homomorphisms are
conjugate in the case where the source group has two generators by studying
centralizer actions in the target group.

##  1 Introduction

Element-conjugate linear representations of finite degree are conjugate if the
underlying field is algebraically closed and of characteristic zero. We are
motivated by exploring the relationship between element-conjugacy and
homomorphism conjugacy for other target groups. M. Larsen initiated this
discussion of the relationship between element-conjugate homomorphisms and
conjugate homomorphisms for Lie groups in his papers from 1994 [1] and 1996
[2]. In the papers, a group $G$ is acceptable if a homomorphism from a group
$\Gamma$ to $G$ is determined up to conjugacy by the conjugacy classes of the
elements $\phi(\gamma)$. Larsen made progress toward the classification of
acceptable subgroups of the general linear group where all algebraic groups
are assumed to be connected even if their associated Lie groups are not.
Weidner later extended this work by using pseudocharacters to answer more
questions about acceptable algebraic groups [3].

We focus on a discrete variation of this problem by determining under which
conditions some generator-conjugate or element-conjugate homomorphisms to the
symmetric group are conjugate.

###  1.1 Notation and Important Definitions

####  1.1.1 Notation

Let $G$ be a group and $x$ be an element in $G$, $\mathrm{Cent}_{G}(x)$
denotes the centralizer of $x$ in $G$. We have that $S_{n}$ is the symmetric
group of the set $\\{1,2,...,n\\}$ while $S_{X}$ is the symmetric group of the
set $X$. Let $\sigma\in S_{n}$, then $\mathbf{Fix}(\sigma)$ is the set of fix
points of $\sigma$ in $\\{1,2,..,n\\}$. We set
$X_{\sigma}=\\{1,2,...,n\\}\setminus\mathbf{Fix}(\sigma)$. Let
$\sigma_{1},\sigma_{2}\in S_{n}$. We set
$X_{\sigma_{1},\sigma_{2}}=\\{1,2,...,n\\}\setminus\mathbf{Fix}(\sigma_{1})\cup\mathbf{Fix}(\sigma_{2}).$

####  1.1.2 Definitions

Let $G$ and $H$ be groups and let $\varphi,\psi$ be homomorphisms from $G$ to
$H$.

######  Definition 1.1 (Generator-conjugate).

Assume $G$ is generated by the elements $g_{1},g_{2},...,g_{n}\in G$ and that
for each $g_{i}$, there exists some $h\in H$ so that

$h\varphi(g_{i})h^{-1}=\psi(g_{i}),$ (1)

then $\varphi,\psi$ are generator-conjugate.

######  Definition 1.2 (Element-conjugate).

If for every $g\in G$, there exists some $h\in H$ so that

$h\varphi(g)h^{-1}=\psi(g),$ (2)

we say $\varphi,\psi$ are element-conjugate.

######  Definition 1.3.

If there exists some $h\in H$ so that for all $g\in G$:

$h\varphi(g)h^{-1}=\psi(g),$ (3)

then $\varphi,\psi$ are conjugate.

###  1.2 Summary

We begin by studying actions induced by centralizers in the symmetric group.
Assume $\sigma\in S_{n}$. The first action is due to the invariance of cycle
type under conjugation in the symmetric group and gives a direct product
decomposition of the centralizer $\mathrm{Cent}_{S_{n}}(\sigma)$(Lemma 2.2).
Let $\sigma^{\prime}\in S_{n}$ be the product of $k$ disjoint cycles of length
$d$,$\;\tau_{1}\tau_{2}\ldots\tau_{k}$. The centralizer of $\sigma^{\prime}$
acts transitively by conjugation on the set
$\\{\tau_{1},\tau_{2},\ldots,\tau_{k}\\}$ (Lemma 2.3).

These two actions to a large extent determine whether two elements
$\pi,\pi^{\prime}\in S_{n}$ are conjugate by some element $\rho$ in the
centralizer of $\sigma$.

Let $\varphi,\psi$ be generator-conjugate homomorphisms from a group $G$ to
the symmetric group. We assume that both homomorphisms send some generator $g$
to the same point $\sigma$ in $S_{n}$. If $\varphi,\psi$ are conjugate as
homomorphisms, there must exist some $\rho\in\mathrm{Cent}_{S_{n}}(\sigma)$ so
that $\rho\varphi\rho^{-1}=\psi$. We are concerned with determining when such
an element $\rho$ exists. Exploring centralizer actions in relation to
homomorphism conjugacy in this way, we obtain our main results.

####  1.2.1 Main results of the paper

Assume $\varphi,\psi$ are homomorphisms from an abelian group with two
generators or a dihedral group, to the symmetric group $S_{n}$. In these cases
we:

  * •

Describe under which conditions generator-conjugate homomorphisms are
conjugate (Theorem 3.1 and Theorem 3.8).

  * •

Detail the relationship between element-conjugate homomorphisms and conjugate
homomorphisms (Corollary 3.6 and Corollary 3.9).

###  1.3 Possible Future Directions

The dihedral case describes the relationship between generator-conjugacy,
element-conjugacy, and conjugacy for homomorphisms from a finite reflection
group with two generators to the symmetric group. A natural next step could be
attempting to generalize some of these results to other finite reflection
groups as source groups. In Lemma 2.3, we prove that the centralizer of an
element $\sigma=\tau_{1}\tau_{2}..\tau_{k}$, the product of $k$-cycles of
length $d$, acts transitively on the set
$\\{\tau_{1},\tau_{2},..,\tau_{k}\\}$. However, we later note that there is
also a larger subgroup $H_{\sigma}$ that acts transitively on the set
$\\{\\{\tau_{1}\\},...\\{\tau_{k}\\}\\}$, where $\\{\tau_{i}\\}$ is the set of
points in the cycle $\tau_{i}$ (Lemma 2.15):

$H_{\sigma}=\\{x\in S_{n}|x\sigma x^{-1}=\sigma^{k},k\in\mathbb{Z}\\}$ (4)

Note that any finite reflection group $S$ can be represented in the form:

$S=\langle
s_{1},s_{2},...,s_{n}\;|\;s_{i}^{2}=1,(s_{i}s_{j})^{m_{i,j}}=1,\;m_{i,j}\in\mathbb{Z}\rangle,$
(5)

and that $s_{i},s_{j}\in H_{s_{i}s_{j}}$:

$s_{i}(s_{i}s_{j})s_{i}=s_{j}(s_{i}s_{j})s_{j}=s_{j}s_{i}=(s_{i}s_{j})^{-1}.$
(6)

A way to work out the relations between element-conjugacy and conjugacy for
homomorphisms to the symmetric group from an arbitrary finite reflection group
might be through this connection to actions, in a similar way to the abelian
or dihedral case we have already covered.

Further, we could also let the target group itself be any finite reflection
group and try to further generalize the results to this setting.

Another potential route is by first noting that $S_{n}$ is the Weyl group of
$\mathrm{GL}(n,\mathbb{C})$. It could then be interesting to explore the
connection between element-conjugacy versus conjugacy for homomorphisms to
each group. This could be expanded into attempting to explain this relation
for other connected, reductive groups $G$ and their corresponding Weyl groups.

##  2 Centralizer Actions in the Symmetric Group

Let $\sigma$ be an arbitrary permutation in $S_{n}$. We begin by defining the
notion of a non-trivial centralizer.

######  Definition 2.1 (Non-trivial Centralizer).

We call the centralizer of $\sigma$ in the symmetric group $S_{X_{\sigma}}$
where $X_{\sigma}=\\{1,2,...,n\\}\setminus\mathbf{Fix}(\sigma)$ the non-
trivial centralizer of $\sigma$ and denote it $\mathrm{Cent}_{0}(\sigma)$.

We may rewrite $\sigma$ as a product of disjoint cycles and group the cycles
of the same length together such that

$\sigma=\sigma_{1}\sigma_{2}...\sigma_{l},$ (7)

where each $\sigma_{i}$ is the product of $k_{i}$ disjoint cycles of length
$d_{i}$ and all the integers $d_{1},...,d_{l}$ are distinct. The first
important centralizer action we observe is induced by the invariance of cycle
type under conjugation in $S_{n}$.

######  Lemma 2.2 (Direct Product Decomposition of Centralizer).

The centralizer of $\sigma=\sigma_{1}\sigma_{2}...\sigma_{l}$ has a direct
product decomposition in $S_{n}$

$\mathrm{Cent}_{S_{n}}(\sigma)=S_{\mathbf{Fix}(\sigma)}\times\mathrm{Cent}_{0}(\sigma_{1})\times\mathrm{Cent}_{0}(\sigma_{2})\times...\times\mathrm{Cent}_{0}(\sigma_{l})$
(8)

###### Proof.

Assume $\pi\in\mathrm{Cent}_{S_{n}}$. Then

$\pi\sigma_{1}\pi^{-1}\pi\sigma_{2}\pi^{-1}...\pi\sigma_{l}\pi^{-1}=\sigma_{1}\sigma_{2}...\sigma_{l}.$
(9)

By invariance of cycle type under conjugation, it must hold that

$\pi\sigma_{i}\pi^{-1}=\sigma_{i},\;\;1\leq i\leq l.$ (10)

Every point in a cycle of length $d_{i}$ in $\sigma_{i}$ must be sent to some
cycle of length $d_{i}$ which also lies in $\sigma_{i}$ by construction. ∎

This direct product decomposition of the centralizer motivates restricting our
further study of centralizer actions in $S_{n}$ to the case where $\sigma$ is
the product of $k$ disjoint cycles of length $d$. We assume $\sigma$ satisfies
this condition for the remainder of this entire section.

The second important action for our results is presented in the following
lemma.

######  Lemma 2.3.

$\mathrm{Cent}_{0}(\sigma)$ acts transitively by conjugation on the set
$\\{\tau_{1},\tau_{2},...,\tau_{k}\\}$.

###### Proof.

The identity element acts trivially on all elements in the set. We show that
the orbit of $\tau_{1}$ is the entire set. We can set $\tau_{1}=(12\cdots d)$
and $\tau_{i}=(i_{1}i_{2}...i_{d})$ and define a permutation:

$\pi(j)=i,\pi(i_{j})=j;1\leq j\leq
d,\;\mathbf{Fix}(\pi)=X_{\sigma}\setminus\\{1,2,...,n\\}\cup\\{i_{1},i_{2},...,i_{d}\\}$
(11)

Clearly, $\pi\in\mathrm{Cent}_{0}(\sigma)$ and $\pi\tau_{1}\pi^{-1}=\tau_{i}$.
Next, assume there exists some element $\rho\in\mathrm{Cent}_{0}(\sigma)$ so
that $\rho\tau_{1}\rho^{-1}\notin\\{\tau_{1},\tau_{2},...,\tau_{k}\\}$. Since
we assume that the cycles $\tau_{1},\tau_{2},..,\tau_{k}$ are disjoint, this
implies that $\rho\tau_{1}...\tau_{k}\rho^{-1}\neq\tau_{1}...\tau_{k}$ which
is a contradiction. ∎

######  Corollary 2.4.

The subgroup $K=\langle\tau_{1},\tau_{2},\tau_{3},...,\tau_{k}\rangle$ is
normal in $\mathrm{Cent}_{0}(\sigma)$.

Note that the associated permutation representation of the action of
$\mathrm{Cent}_{0}(\sigma)$ on $\\{\tau_{1},...,\tau_{k}\\}$ is precisely
given by the homomorphism whose kernel is $K$:

$\mathrm{Cent}_{0}(\sigma)\longrightarrow\mathrm{Cent}_{0}(\sigma)/K\cong
S_{k}$ (12)

Each element $\pi\in\mathrm{Cent}_{0}(\sigma)$ may therefor by its image under
$\mathrm{Cent}_{0}(\sigma)$ be associated with some element in $S_{k}$,
describing its action on the cycles $\tau_{1},...,\tau_{k}$.

######  Remark 2.5.

Let $\pi\in\mathrm{Cent_{0}}(\sigma)$. The cycle type of $\pi$ under the
homomorphism $\mathrm{Cent}_{0}(\sigma)/K$ is invariant under conjugation in
$\mathrm{Cent}_{0}(\sigma)$.

This is just the result that cycle type is invariant under conjugation in the
symmetric group. However, this remark is significant for our discussion about
homomorphism conjugacy since two elements
$\pi,\pi^{\prime}\in\mathrm{Cent}_{0}(\sigma)$ may be conjugate in $S_{n}$ but
have distinct cycle types under $\mathrm{Cent}_{0}(\sigma)/K$.

######  Example 2.6.

Assume $\sigma=(12)(34)(56)(78)$ and set $\pi=(1324)(5768)$ and
$\pi^{\prime}=(1357)(2468)$. The image of $\pi$ under
$\mathrm{Cent}_{0}(\sigma)/K$ is $(12)(34)$ while the image of $\pi^{\prime}$
under $\mathrm{Cent}_{0}(\sigma)$ is $(1234)$

We now split our focus on centralizer actions in the symmetric group in two
separate directions, based on the two source groups (abelian or dihedral) we
are interested in exploring homomorphism conjugacy for.

In the abelian case, we would like to determine when for two elements
$\pi,\pi^{\prime}\in\mathrm{Cent}_{0}(\sigma)$, there exists some
$\rho\in\mathrm{Cent}_{0}(\sigma)$ so that

$\rho\pi^{\prime}\rho^{-1}=\pi.$ (13)

In the dihedral case, we are interested in elements $\pi,\pi^{\prime}$,
conjugate in $S_{n}$, where

$\pi^{2}=(\pi^{\prime})^{2}=1\;\;\mathrm{and}\;\;\pi\sigma\pi^{-1}=(\pi^{\prime})\sigma(\pi^{\prime})^{-1}=\sigma^{-1}.$
(14)

In particular, we wish to determine when there exists some
$\rho\in\mathrm{Cent_{0}}(\sigma)$ such that

$\rho\pi^{\prime}\rho^{-1}=\pi.$ (15)

###  2.1 Centralizer Actions for Abelian Source

Assume as before that $\sigma\in S_{n}$ is the product of $k$ disjoint cycles
of length $d$:

$\tau_{1}\tau_{2}...\tau_{k}.$ (16)

From now on, we also assume that $\mathbf{Fix}(\sigma)=\emptyset$. The primary
goal of this subsection is to completely determine under which conditions two
elements $\pi,\pi^{\prime}\in\mathrm{Cent}_{0}(\sigma)$ are conjugate in
$\mathrm{Cent}_{0}(\sigma)$. It is sufficient to explore this problem for the
case where the image of $\pi$ is a single cycle of length $m$ under
$\mathrm{Cent}_{0}(\sigma)/K$:

$\mathrm{Cent}_{0}(\sigma)/K:\pi\mapsto(12\cdots m).$ (17)

######  Remark 2.7.

If $\pi,\pi^{\prime}$ are conjugate in $\mathrm{Cent}_{0}(\sigma)$, they have
the same cycle type under $\mathrm{Cent}_{0}(\sigma)/K$.

We have that this follows immediately from the invariance of cycle type under
conjugation in the symmetric group. This in conjunction with that
$K=\langle\tau_{1},\tau_{2},..,\tau_{k}\rangle$ is normal in
$\mathrm{Cent}_{0}(\sigma)$ by Corollary 2.4 means we may write any element in
the non-trivial centralizer as a product of some element in $K$ and some
element not in $K$, whose preimage of the identity under
$\mathrm{Cent}_{0}(\sigma)/K$ is trivial.

######  Remark 2.8.

We can rewrite the elements $\pi,\pi^{\prime}$ such that

$\pi=\mu\pi_{0},\pi=\mu^{\prime}\pi_{0}^{\prime},$ (18)

where $\mu,\mu^{\prime}\in K$ and $\pi_{0},\pi_{0}^{\prime}$ permutes no
cycles trivially, i.e either all points in a cycle $\tau_{i}$ are fixed or
sent to the points in some distinct cycle $\tau_{j}$.

This in conjunction with the action on the set $\\{\tau_{1},...,\tau_{k}\\}$
(Lemma 2.3) results in an additional condition for conjugacy in
$\mathrm{Cent}_{0}(\sigma)$:

######  Remark 2.9.

There are integers $z_{1},z_{2},...,z_{k}$ and $n_{1},n_{2},...,n_{k}$ such
that

$\mu=\tau_{1}^{z_{1}}\tau_{2}^{z_{2}}...\tau_{k}^{z_{k}}\;\;\mathrm{and}\;\;\mu^{\prime}=\tau_{1}^{n_{1}}\tau_{2}^{n_{2}}...\tau_{k}^{n_{k}}.$
(19)

If $\pi,\pi^{\prime}$ are conjugate in $\mathrm{Cent}_{0}(\sigma)$, then the
lists of integers $z_{1},z_{2},...,z_{k}$ and $n_{1},n_{2},...,n_{k}$ are
identical up to rearrangement.

We now make an important observation about $\pi_{0}$ in the decomposition
$\pi=\mu\pi_{0}$ as in Remark 2.8.

######  Lemma 2.10.

We have that

$\pi_{0}^{m}=\tau_{1}^{z}\tau_{2}^{z}...\tau_{m}^{z},$ (20)

for some integer $z$.

###### Proof.

It is clear that $\pi_{0}^{m}\in\ker(\mathrm{Cent}_{0}(\sigma)/K)$. Then
$\pi_{0}^{m}=\tau_{1}^{z_{1}}\tau_{2}^{z_{2}}...\tau_{m}^{z_{m}}$ for some
integers $z_{1},z_{2},..,z_{m}$. Assume there exists a pair of integers $i<j$
such that $z_{i}\neq z_{j}$. This implies that $\pi_{0}^{j-i}$ does not
commute with $\pi_{0}^{m}$ which is a contradiction. ∎

This lemma makes it possible to define the last necessary condition for
conjugacy in $\mathrm{Cent}_{0}(\sigma)$.

######  Remark 2.11.

We rewrite the elements $\pi,\pi^{\prime}$ such that

$\pi=\mu\pi_{0},\pi=\mu^{\prime}\pi_{0}^{\prime},$ (21)

as in Remark 2.8. If $\pi,\pi^{\prime}$ are conjugate in
$\mathrm{Cent}_{0}(\sigma)$, then
$\pi_{0}^{m}=\tau_{1}^{z}\tau_{2}^{z}...\tau_{m}^{z}$ and
$(\pi_{0}^{\prime})^{m}=\tau_{i_{1}}^{z}\tau_{i_{2}}^{z}...\tau_{i_{m}}^{z}$
for an integer $z$.

###### Proof.

By that $K$ is normal in $\mathrm{Cent}_{0}(\sigma)$, we must have that
$\pi_{0},\pi_{0}^{\prime}$ are conjugate in $\mathrm{Cent}_{0}(\sigma)$.
Assume that $\pi_{0}^{m}=\tau_{1}^{z}...$ and
$(\pi_{0}^{\prime})^{m}=\tau_{i_{1}}^{z^{\prime}}...$ for some distinct
integers $z,z^{\prime}$. Then $\tau_{1}^{z},\tau_{i_{1}}^{z^{\prime}}$ are
conjugate in $\mathrm{Cent}_{0}(\sigma)$. This implies that the orbit of
$\tau_{1}$ under the action of $\mathrm{Cent}_{0}(\sigma)$ is not the set
$\\{\tau_{1},\tau_{2},..,\tau_{k}\\}$ which is a contradiction by Lemma 2.3. ∎

We are ready to define precisely when $\pi,\pi^{\prime}$ are conjugate in
$\mathrm{Cent}_{0}(\sigma)$.

######  Lemma 2.12.

Let $\pi,\pi^{\prime}$ be two elements in the non-trivial centralizer of
$\sigma=\tau_{1}...\tau_{k}$ where

$\mathrm{Cent}_{0}(\sigma)/K:\pi\mapsto(12\cdots m),$ (22)

recalling that $K=\langle\tau_{1},\tau_{2},...,\tau_{k}\rangle$. We may
rewrite $\pi,\pi^{\prime}$ by Remark 2.8 so that $\pi=\mu\pi_{0}$ and
$\pi^{\prime}=\mu^{\prime}\pi_{0}^{\prime}$ where $\mu,\mu^{\prime}\in K$ and
$\pi_{0},\pi_{0}^{\prime}\notin K$. Then
$\mu_{0}=\tau_{1}^{z_{1}}...\tau_{k}^{z_{k}}$ for some integers
$z_{1},z_{2},..,z_{k}$ and $\mu_{0}=\tau_{1}^{n_{1}}...\tau_{k}^{n_{k}}$ for
some integers $n_{1},n_{2},..,n_{k}$. The elements $\pi,\pi^{\prime}$ are
conjugate in $\mathrm{Cent}_{0}(\sigma)$ if and only if:

  1. (i)

The image of $\pi^{\prime}$ under $\mathrm{Cent}_{0}(\sigma)/K$ is also an
$m$-cycle:

$\mathrm{Cent}_{0}/K:\pi^{\prime}=(i_{1}i_{2}\cdots i_{m}).$ (23)

  2. (ii)

The lists $z_{1},..,z_{k}$ and $n_{1},...,n_{k}$ are identical up to
rearrangement.

  3. (iii)

There exists an integer $z$ so that $\pi_{0}^{m}=\tau_{1}^{z}...\tau_{m}^{z}$
and $(\pi_{0}^{\prime})^{m}=\tau_{i_{1}}^{z}...\tau_{i_{m}}^{z}$

###### Proof.

It follows from Remark 2.7, 2.9, and 2.11 that each of the conditions are
necessary. It remains to prove that they are sufficient. Look at the list of
integers $n_{1},..,n_{k}$. It is clear that

$n_{i_{1}}=n_{i_{2}}=...=n_{i_{m}}=0,$ (24)

by the image of $\pi^{\prime}$ under $\mathrm{Cent}_{0}(\sigma)/K$. By the
transitive action on the cycles $\\{\tau_{1},\tau_{2},..,\tau_{k}\\}$ and the
fact that the lists $z_{1},..,z_{k}$ and $n_{1},...,n_{k}$ are identical up to
rearrangement there exists some $\rho\in\mathrm{Cent}_{0}(\sigma)$ such that

$\rho\mu_{0}^{\prime}\rho^{-1}=\mu_{0}\;\;\mathrm{and}\;\;\mathrm{Cent_{0}(\sigma)}/K:\rho\pi^{\prime}\rho^{-1}\mapsto(12\cdots
m).$ (25)

We are done if we can prove that there exists some integers
$j_{1},j_{2},...,j_{m}$ so that

$(\tau_{1}^{j_{1}}\tau_{2}^{j_{2}}...\tau_{m}^{j_{m}})(\rho\pi_{0}^{\prime}\rho^{-1})(\tau_{1}^{j_{1}}\tau_{2}^{j_{2}}...\tau_{m}^{j_{m}})^{-1}=\pi_{0}.$
(26)

Set $\rho\pi_{0}^{\prime}\rho^{-1}=\pi_{0}^{\prime\prime}$. By condition
(iii), there exists some integer $z$ so that

$(\pi_{0})^{m}=(\pi_{0}^{\prime\prime})^{m}=\tau_{1}^{z}\tau_{2}^{z}...\tau_{m}^{z}.$
(27)

This implies that $\pi_{0},\pi_{0}^{\prime\prime}$ have the same cycle type.
Let $d^{\prime}$ denote the length of the cycles in $\tau_{1}^{z}$. Then
$\pi_{0},\pi_{0}^{\prime\prime}$ have cycle type $\frac{d}{d^{\prime}}$
disjoint cycles of length $md^{\prime}$.

We observe that the elements $\pi_{0},\pi_{0}^{\prime\prime}$ are completely
determined by one of their cycles. To see this, let $c_{0}$ denote a cycle in
$\pi_{0}$ and $c_{0}^{\prime\prime}$ denote a cycle in
$\pi_{0}^{\prime\prime}$. Then

$\begin{split}c_{0}&=(x_{\tau_{1}}\cdots x_{\tau_{m}}x_{\tau_{1}+z}\cdots
x_{\tau_{m}+z}\cdots x_{\tau_{1}+d^{\prime}z}\cdots
x_{\tau_{m}+d^{\prime}z})\\\ c_{0}^{\prime\prime}&=(y_{\tau_{1}}\cdots
y_{\tau_{m}}y_{\tau_{1}+z}\cdots y_{\tau_{m}+z}\cdots
y_{\tau_{1}+d^{\prime}z}\cdots y_{\tau_{m}+d^{\prime}z}),\end{split}$ (28)

where $x_{\tau_{i}},y_{\tau_{i}}$ are points in the cycles $\tau_{i}$ ,
$\tau_{i}^{hz}(x_{\tau_{i}})=x_{\tau_{i}+hz}$, and
$\tau_{i}^{hz}(y_{\tau_{i}})=y_{\tau_{i}+hz}$. These relations completely
determine how the remaining points in the cycles $\tau_{1},...\tau_{m}$ are
sent. We see that $\pi,\pi^{\prime\prime}$ are not only determined by the
cycles $c_{0},c_{0}^{\prime\prime}$ but further, completely determined by the
first $m$ points in the cycles. Naturally, there exists integers
$h_{1},..,h_{m}$ so that

$\tau_{i}^{h_{i}}(x_{\tau_{i}})=y_{\tau_{i}},\;\;1\leq i\leq m,$ (29)

which concludes our argument. ∎

######  Corollary 2.13.

Assume that the order of $\sigma$, $|\tau_{1}...\tau_{k}|=d$ and the order of
$\pi$ are relatively prime, then $\pi$ and $\pi^{\prime}$ are conjugate if
$\pi,\pi^{\prime}$ are conjugate in $S_{n}$.

###### Proof.

By $\pi,\pi^{\prime}$ having order relatively prime to $d$, we must have that
$\mu=\mu^{\prime}=1$. Also $\pi^{m}=1$ and since $\pi^{\prime}$ conjugate to
$\pi$ and relatively prime to $d$ , the image of $\pi^{\prime}$ is also an
$m$-cycle and $(\pi^{\prime})^{m}=1$. The result follows from Lemma 2.12. ∎

######  Corollary 2.14.

Assume that the order $\sigma$, $d=2$. The elements $\pi,\pi^{\prime}$ are
conjugate in $\mathrm{Cent}_{0}(\sigma)$ if $\pi,\pi^{\prime}$ are conjugate
in $S_{n}$ and have the same cycle type under $\mathrm{Cent}_{0}(\sigma)/K$

###### Proof.

If $\pi,\pi^{\prime}$ have odd order, the result follows from the corollary
above. Assume $\pi,\pi^{\prime}$ have even order.
$\pi^{\prime},\pi_{0}^{\prime}$ act non-trivially on $md$ points. It is clear
that $\pi,\pi^{\prime}$ have the same number of fix points and so we can
assume that there exists $\rho\in\mathrm{Cent}_{0}(\sigma)$ so that
$\rho\mu^{\prime}\rho^{-1}=\mu$ and $\rho\pi_{0}^{\prime}\rho^{-1},\pi$ have
the same image under $\mathrm{Cent}_{0}(\sigma)/K$. The conjugacy in $S_{n}$
further forces that $(\pi_{0}^{\prime})^{m}=(\pi_{0})^{m}=1$ or
$(\pi_{0}^{\prime})^{m}=(\pi_{0})^{m}=\tau_{1}...\tau_{k}$. The result now
follows from 2.12. ∎

Note that Corollary 2.13 and 2.14 can easily be generalized to the case where
$\pi$ has arbitrary cycle type under $\mathrm{Cent}_{0}(\sigma)/K$.

###  2.2 Centralizer Actions for Dihedral Source

Assume as before that $\sigma\in S_{n}$ is the product of $k$ disjoint cycles
of length $d$, $\tau_{1}\tau_{2}...\tau_{k}$ and
$\mathbf{Fix}(\sigma)=\emptyset$. We are interested in exploring elements
$\pi,\pi^{\prime}\in S_{n}$ of the type

$\pi^{2}=(\pi^{\prime})^{2}=1\;\;\mathrm{and}\;\;\pi\sigma\pi^{-1}=(\pi^{\prime})\sigma(\pi^{\prime})^{-1}=\sigma^{-1}.$
(30)

We start by noting that $\pi,\pi^{\prime}$ are elements in a larger subgroup
than $\mathrm{Cent}_{0}(\sigma)$, $H_{\sigma}$, where we also have a
transitive action on the cycles $\\{\tau_{1},\tau_{2},...,\tau_{k}\\}$:

######  Lemma 2.15.

The subgroup $H_{\sigma}=\\{x\in S_{n}|x\sigma
x^{-1}=\sigma^{z},\;z\in\mathbb{Z}\\}$ acts transitively on the cycles
$\\{\tau_{1},\tau_{2},...,\tau_{k}\\}$. The kernel $K_{H_{\sigma}}$ of this
action is

$K_{H_{\sigma}}=\\{x\in
H_{\sigma}|x\tau_{i}x^{-1}=\tau_{i}^{z},\;\;z\in\mathbb{Z}\;\mathrm{for}\;\mathrm{all}\;1\leq
i\leq k\\}.$ (31)

###### Proof.

This is straightforward to prove, see Lemma 2.3. ∎

It is obvious that $\mathrm{Cent}_{0}(\sigma)\leq H_{\sigma}$, and in fact
$\mathrm{Cent}_{0}(\sigma)\unlhd H_{\sigma}.$ As for the non-trivial
centralizer, we get a permutation representation, this time from $H_{\sigma}$
to $S_{k}$:

$H_{\sigma}\to H_{\sigma}/K_{H_{\sigma}}\cong S_{k}.$ (32)

The main goal of this subsection is to prove that if $\pi,\pi^{\prime}$ are
conjugate in $S_{n}$ and if the images of the elements under
$H_{\sigma}/K_{H_{\sigma}}$ are also conjugate in $S_{k}$, then there exists
some $\rho\in\mathrm{Cent}_{0}(\sigma)$ so that

$\rho(\pi^{\prime})\rho^{-1}=\pi.$ (33)

We first look at elements $s_{1},s_{2}\in S_{X_{\tau_{1}}}$, that send a
single cycle $\tau_{1}$ to its own inverse or elements $s_{1},s_{2}\in
S_{X_{\tau_{1},\tau_{2}}}$ that sends the cycle $\tau_{1}$ to the inverse of a
distint cycle $\tau_{2}$.

######  Lemma 2.16.

Assume $\tau_{1}=(12\cdots d)$ and $\tau_{2}=(j_{1}j_{2}\cdots j_{d})$. Let
$s_{1},s_{2}\in S_{X_{\tau_{1},\tau_{2}}}$ both be products of disjoint
transpositions such that

$s_{1}\tau_{1}s_{1}^{-1}=s_{2}\tau_{1}s_{2}^{-1}=\tau_{2}^{-1}.$ (34)

There exists an integer $z$ such that

$\tau_{2}^{z}s_{1}\tau_{2}^{-z}=s_{2}.$ (35)

###### Proof.

There are precisely $d$ ways to send the cycle $\tau_{1}$ to $\tau_{j}^{-1}$
through transpositions. Once we send the point 1 in $\tau_{1}$ to some point
in $\tau_{2}$, the element is determined. Assume $s_{1}=(1j_{1})...$ and
$s_{2}=(1j_{i})...$. Then

$\tau_{2}^{i-1}s_{1}\tau_{2}^{-(i-1)}=s_{2}.$ (36)

∎

######  Lemma 2.17.

Let $\tau_{1}=(12\cdots d)$ and let $s_{1},s_{2}\in S_{X_{\tau_{1}}}$ be
products of disjoint transpositions such that

$s_{1}\tau_{1}s_{1}^{-1}=s_{2}\tau_{1}s_{2}^{-1}=\tau_{1}^{-1}.$ (37)

If $s_{1},s_{2}$ are conjugate in $S_{n}$, there exists some integer $z$ so
that:

$\tau_{1}^{z}s_{1}\tau_{1}^{-z}=s_{2}.$ (38)

###### Proof.

The elements $s_{1},s_{2}$ may precisely be associated with reflections along
the lines of symmetry of a regular polygon with $d$ vertices. If $d$ is odd,
$s_{1},s_{2}$ will have cycle type $\frac{d-1}{2}$ transpositions. The lines
of symmetry of a regular polygon with an odd number of vertices is completely
determined by the vertex it intersects. Assume $s_{1}$ fixes vertex 1 and
$s_{2}$ fixes vertex $i$. Then $\tau_{1}^{i-1}s_{1}\tau_{1}^{-(i-1)}=s_{2}$.

If $d$ is even, a line of symmetry either intersects two vertices or an edge.
In the former case $s_{1},s_{2}$ have cycle type $\frac{d-2}{2}$ and the
elements are again completely determined by a fixed vertex so that if $s_{1}$
fixes vertex 1 and $s_{2}$ fixes vertex i, then
$\tau_{1}^{i-1}s_{1}\tau_{1}^{-(i-1)}=s_{2}$.

In the latter case, $s_{1},s_{2}$ have cycle type $\frac{d}{2}$ transpositions
and the elements are determined by an edge that the line of symmetry
intersects. Let $s_{1}=(12)...$ and $s_{2}=(i\;i+1)...$ , then
$\tau_{1}^{i-1}s_{1}\tau_{1}^{-(i-1)}=s_{2}$ and we are done. ∎

We are ready to state the main lemma:

######  Lemma 2.18.

Let $\pi,\pi^{\prime}$ be elements conjugate in $S_{n}$ and assume that their
images under $H_{\sigma}/K_{H_{\sigma}}$ have the same cycle type. Then, there
exists some $\rho\in\mathrm{Cent}_{0}(\sigma)$ so that
$\rho\pi\rho^{-1}=\pi^{\prime}$.

###### Proof.

We may write $\pi,\pi^{\prime}$ in the form $\pi=\mu\pi_{0}$,
$\pi^{\prime}=\mu^{\prime}\pi_{0}^{\prime}$ where $\mu,\mu^{\prime}\in
K_{H_{\sigma}}$ and $\pi_{0},\pi_{0}^{\prime}$ either fix all points in a
cycle $\tau_{i}$ or sends all the points to a distinct cycle $\tau_{j}$.
Assume $\mu$ acts non-trivially on the points in the cycles
$\tau_{1},...,\tau_{m}$. Then, $\pi_{0}$ acts non-trivially on the cycles
$\tau_{m+1},...,\tau_{k}$. Since $\pi,\pi^{\prime}$ have the same cycle type
under $H_{\sigma}/K_{H_{\sigma}}$, this implies that $\mu,\mu^{\prime}$ are
conjugate in $S_{n}$. By the transitive action on the cycles
$\tau_{1},..,\tau_{k}$ in $\mathrm{Cent}_{0}(\sigma)$, this in turn implies
that there exists some $\rho\in\mathrm{Cent}_{0}(\sigma)$ so that
$\rho\mu^{\prime}\rho^{-1}$ acts non-trivially on the cycles
$\tau_{1},...,\tau_{m}$ and $\rho\pi_{0}^{\prime}\rho^{-1},\rho\pi\rho^{-1}$
have the same image under $H_{\sigma}/K_{H_{\sigma}}$. The result now follows
from Lemma 2.16 and 2.17. ∎

######  Corollary 2.19.

If the order of $\sigma$, $|\tau_{1}...\tau_{k}|=d$ is odd and
$\pi,\pi^{\prime}$ conjugate in $S_{n}$, there exists some
$\rho\in\mathrm{Cent}_{0}(\sigma)$ so that $\rho\pi\rho^{-1}=\pi^{\prime}$

###### Proof.

If $d$ is odd and $\pi,\pi^{\prime}$ conjugate in $S_{n}$, this forces
$\pi,\pi^{\prime}$ to have the same image under $H_{\sigma}/K_{H_{\sigma}}$.
The result follows from Lemma 2.18. ∎

##  3 Homomorphism Conjugacy in the Symmetric Group

Let $\varphi,\psi$ be two generator-conjugate homomorphisms from a group $G$
to $S_{n}$ with respect to the generators $g_{1},....,g_{n}$. We first wish to
determine when $\varphi,\psi$ are conjugate as homomorphisms. We may assume
that

$\varphi(g_{1})=\psi(g_{1}),$ (39)

so that if $\varphi,\psi$ are conjugate, there exists some
$\rho\in\mathrm{Cent}_{S_{n}}(\varphi(g_{1}))$ so that

$\rho\varphi\rho^{-1}=\psi.$ (40)

This observation, in conjunction with the results derived about centralizer
actions from the previous section, gives the first part of our main result:
determining when generator-conjugate homomorphisms to $S_{n}$ are conjugate
for an abelian or dihedral source group with two generators. The second part
of the main result, describing the relationship between element-conjugate and
conjugate homomorphisms for such source groups, is derived by first studying
the relationship between generator-conjugate and element-conjugate
homomorphisms through centralizer actions in $S_{n}$.

###  3.1 Homomorphisms Conjugacy for Abelian Source

Let $\varphi,\psi$ be two homomorphisms from an abelian group $A$, generated
by two elements $a,b$ to the symmetric group $S_{n}$.

Assume $\varphi,\psi$ are generator-conjugate with respect to $a,b$ and that

$\varphi(a)=\psi(a).$ (41)

We may set $\varphi(a)=\sigma$ and rewrite the element so that
$\sigma=\sigma_{1}...\sigma_{l}$ where $\sigma_{i}$ is a product of $k_{i}$
disjoint cycles of length $d_{i}$,
$\tau_{(1,i)}\tau_{(2,i)}...\tau_{(k_{i},i)}$, and the integers
$d_{1},...,d_{l}$ are all distinct. By the direct product decomposition of
centralizers in the symmetric group (Lemma 2.2), we may also rewrite
$\varphi(b),\psi(b)$:

$\begin{split}\varphi(b)&=\mu_{0}\pi_{1}\pi_{2}...\pi_{l}\\\
\psi(b)&=\mu_{0}^{\prime}\pi_{1}^{\prime}\pi_{2}^{\prime}...\pi_{l}^{\prime},\end{split}$
(42)

where $\mu_{0},\mu_{0}^{\prime}\in S_{\mathbf{Fix(\sigma}}$ and
$\pi_{i},\pi_{i}^{\prime}\in\mathrm{Cent}_{0}(\sigma_{i})$. By Remark 2.8,
each $\pi_{i},\pi_{i}^{\prime}$ may decomposed as products:
$\pi_{i}=\mu_{i}\pi_{0,i}$ and $\pi_{i}^{\prime}=\mu_{i}\pi_{0,i}^{\prime}$
where $\mu_{i},\mu_{i}^{\prime}\in
K_{i}=\langle\tau_{1,i},...,\tau_{k_{i},i}\rangle$ and
$\pi_{i},\pi_{i}^{\prime}$ either fix all points in a cycle $\tau_{j,i}$ or
sends all points to some distinct cycle $\tau_{j^{\prime},i}$. Note that we
have that

$\begin{split}\mu_{i}&=\tau_{1,i}^{z_{(1,i)}}\tau_{2,i}^{z_{(2,i)}}...\tau_{k_{i},i}^{z_{(k_{i},i)}}\\\
\mu_{i}^{\prime}&=\tau_{1,i}^{z_{(1,i)}^{\prime}}\tau_{2,i}^{z_{(2,i)}^{\prime}}...\tau_{k_{i},i}^{z_{(k_{i},i)}^{\prime}},\end{split}$
(43)

for some integers $z_{(1,i)},..,z_{(k_{i},i)}$ and
$z_{(1,i)}^{\prime},..,z_{(k_{i},i)}^{\prime}$.

The action in $\mathrm{Cent}_{0}(\sigma_{i})$ gives a further decomposition of
the elements $\pi_{(0,i)}$ and $\pi_{(0,i)}^{\prime}$ as products of elements
in the non-trivial centralizer:

$\begin{split}\pi_{(0,i)}&=x_{(1,i)}....x_{(h,i)}\\\
\pi_{(0,i)}^{\prime}&=y_{(1,1)}....y_{(h^{\prime},i)},\end{split}$ (44)

where the image of every $x_{(j,i)},y_{(j,i)}$ is a single cycle under
$\mathrm{Cent}_{0}(\sigma_{i})/K_{i}$.

######  Theorem 3.1.

The homomorphisms $\varphi,\psi$ are conjugate if and only if:

  1. (i)

$\mu_{0},\mu_{0}^{\prime}$ are conjugate in $S_{n}$.

  2. (ii)

The lists $z_{(1,i)},..,z_{(k_{i},i)}$ and
$z_{(1,i)}^{\prime},..,z_{(k_{i},i)}^{\prime}$are identical up to
rearrangement for $1\leq i\leq l$ .

  3. (iii)

For every $x_{(j,i)}$ whose image is an $m$-cycle under
$\mathrm{Cent}_{0}(\sigma_{i})/K_{i}$ there exists some $y_{(j^{\prime},i)}$
whose image is also an $m$-cycle under $\mathrm{Cent}_{0}(\sigma_{i})/K_{i}$
where $x_{(j,i)}^{m}=\tau_{t,i}^{z_{j,i}}...$ and
$y_{j^{\prime}}^{m}=\tau_{t^{\prime},i}^{z_{j,i}}...$ for some integer
$z_{j,i}$.

###### Proof.

By the direct product decomposition of the centralizer of
$\varphi(a)=\psi(a)=\sigma$ we must assume that there exists some $\rho_{0}\in
S_{\mathbf{Fix}(\sigma)}$ so that $\rho_{0}\mu_{0}^{\prime}\rho_{0}=\mu_{0}$
and $\rho_{i}\in\mathrm{Cent}_{0}(\sigma_{i})$ so that
$\rho_{i}\pi_{i}^{\prime}\rho_{i}=\pi_{i}$ for $1\leq i\leq l$. Condition (i)
follows from that cycle type determines conjugacy in symmetric groups.
Condition (ii) follows directly from Lemma 2.12. Condition (iii) follows from
the decomposition of $\pi_{0,i},\pi_{0,i}^{\prime}$ into disjoint elements
whose image under $\mathrm{Cent}_{0}(\sigma_{i})/K_{i}$ is a single cycle and
Lemma 2.12. ∎

The following two corollaries follow quite easily from Theorem 3.1 and
generalizing the results in Corollary 2.13 and 2.14.

######  Corollary 3.2.

If the order of $a,b$ are relatively prime then $\varphi,\psi$ are conjugate
if and only if: $\mu_{0},\mu_{0}^{\prime}$ conjugate in $S_{n}$ and
$\pi_{i},\pi_{i}^{\prime}$ are conjugate in $S_{n}$ for $1\leq i\leq l$.

######  Corollary 3.3.

If $|a|=2$ then $\varphi,\psi$ are conjugate if and only if:
$\mu_{0},\mu_{0}^{\prime}$ conjugate in $S_{n}$, $\pi_{i},\pi_{i}^{\prime}$
are conjugate in $S_{n}$, and the elements also have the same cycle type under
$\mathrm{Cent}_{0}(\sigma_{i})/K_{i}$ for $1\leq i\leq l$.

Note that although the source group is cyclic in Corollary 3.2, the result is
not trivial. We can easily define generator-conjugate homomorphisms from
cyclic source groups that are not conjugate.

######  Example 3.4.

Consider the homomorphisms
$\varphi,\psi:A\cong\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\to
S_{12}:$

$\begin{split}&\varphi,\psi:a\mapsto(123)(456);\\\
&\varphi:b\mapsto(14)(25)(36),\;\psi:b\mapsto(78)(9\;10)(11\;12).\end{split}$
(45)

The homomorphisms are clearly generator-conjugate, but one can check that
$\varphi(ab),\psi(ab)$ have different cycle types in $S_{n}$.

Further, the conditions set out in Theorem 3.1 make it easy to define distinct
generator-conjugate homomorphisms from a non-cyclic abelian group to $S_{n}$.

######  Example 3.5.

Consider the homomorphisms
$\varphi,\psi:A\cong\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\to
S_{11}:$

$\begin{split}&\varphi,\psi:a\mapsto(1234)(5678);\\\
&\varphi:b\mapsto(15)(26)(37)(48)(9\;10),\;\psi:b\mapsto(16)(27)(38)(45)(10\;11).\end{split}$
(46)

These homomorphisms are conjugate by Theorem 3.1 and it is straightforward to
check they are distinct.

We conclude this subsection by describing the relationship between element-
conjugate and conjugate homomorphisms:

######  Corollary 3.6.

The homomorphisms $\varphi,\psi$ are conjugate if and only if they are
element-conjugate on restriction to $S_{\mathrm{Fix}(\sigma)}$ and
$\mathrm{Cent}_{0}(\sigma_{i})$, for $1\leq i\leq l$.

If $\varphi,\psi$ are conjugate as homomorphisms, it is clear by the direct
product decomposition of the centralizer that we must assume that the
homomorphisms are also element-conjugate when restricted to $S_{Fix(\sigma)}$
and each subgroup $\mathrm{Cent}_{0}(\sigma_{i})$. It is then relevant to
restrict ourselves to investigating when generator-conjugate homomorphisms are
element-conjugate in the case where $\psi(a)=\psi(a)$ is the product of $k$
disjoint cycles of length $d$, $\tau_{1}...\tau_{k}$ and this element has no
fix points in $S_{n}$. Recall that in this case, we can decompose
$\varphi(b),\psi(b)$ such that:

$\begin{split}\varphi(b)&=\mu x_{1}....x_{h}\\\
\psi(b)&=\mu^{\prime}y_{1}....y_{h^{\prime}},\end{split}$ (47)

where $\mu,\mu^{\prime}\in K=\langle\tau_{1},...,\tau_{k}\rangle$ and the
images of all $x_{j},y_{j}$ under $\mathrm{Cent}_{0}(\sigma)/K$ are single
cycles and $x_{i},y_{i}$ do not non-trivially send points in a cycle to
itself. We also have that $\mu=\tau_{1}^{z_{1}}...\tau_{k}^{z_{k}}$ and
$\mu^{\prime}=\tau_{1}^{n_{1}}...\tau_{k}^{n_{k}}$ for some integers
$z_{1},...,z_{k}$ and $n_{1},...,n_{k}$.

We present an important lemma that relates generator-conjugacy to element-
conjugacy:

######  Lemma 3.7.

Assume $\psi(a)=\varphi(a)=\tau_{1}....\tau_{k}$ and also that
$\mathbf{Fix}(\tau_{1}...\tau_{k})=\emptyset$. If $\varphi,\psi$ are
generator-conjugate, they are element-conjugate if and only if:

  1. (i)

The lists $z_{i},..,z_{k}$ and $n_{1},..,n_{k}$ are identical up to
rearrangement.

  2. (ii)

For every $x_{j}$ whose image is an $m$-cycle under
$\mathrm{Cent}_{0}(\tau_{1}...\tau_{k})/K$ there exists some $y_{j^{\prime}}$
whose image is also an $m$-cycle under $\mathrm{Cent}_{0}(\sigma_{i})/K_{i}$
and $x_{j,}^{m}=\tau_{i}^{z}...$,
$y_{j^{\prime}}^{m}=\tau_{i^{\prime}}^{z}...$ for some integer $z$.

###### Proof.

If $\varphi,\psi$ are element-conjugate, $\varphi(a^{h_{1}}b^{h_{2}})$ and
$\psi(a^{h_{1}}b^{h_{2}})$ have the same number of fix points for all integers
$h_{1},h_{2}$. This immediately implies that the lists $z_{i},..,z_{k}$ and
$n_{1},..,n_{k}$ are identical up to rearrangement. Ordering the elements
$x_{1},...,x_{h}$ after their cycle length under $\mathrm{Cent}_{0}(\sigma)$,
$m_{1}\leq m_{2}\leq...\leq m_{h}$ , the rest of the proof follows from
comparing fix points and a simple induction argument. ∎

Note that this lemma is precisely the conditions required for homomorphism
conjugacy in this case, and Corollary 3.6 follows.

###  3.2 Homomorphism Conjugacy for Dihedral Source

Let $\varphi,\psi$ be homomorphisms from the dihedral group $D_{2m}$ to
$S_{n}$ and assume the homomorphisms are generator-conjugate with respect to
the generators $r,s$:

$D_{2m}=\langle r,s\;|\;r^{m}=s^{2}=1\;srs=r^{-1}\rangle.$ (48)

We assume $\varphi(r)=\psi(r)=\sigma$ and rewrite $\sigma$ in the form
$\sigma=\sigma_{1}...\sigma_{l}$ where $\sigma_{i}$ is the product $k_{i}$
disjoint cycles of length $d_{i}$ and all the integers $d_{1},..,d_{l}$ are
distinct. By the invariance of cycle type under conjugation, the elements
$\varphi(s),\psi(s)$ can be written as products:

$\begin{split}\varphi(s)&=\mu\pi_{1}....\pi_{l}\\\
\psi(s)&=\mu^{\prime}\pi_{1}^{\prime}....\pi_{l}^{\prime},\end{split}$ (49)

where $\mu,\mu^{\prime}\in S_{\mathbf{Fix}(\sigma)}$ and
$\pi_{i},\pi_{i}^{\prime}\in H_{\sigma_{i}}$ recalling that

$H_{\sigma_{i}}=\\{x\in
S_{X_{\sigma_{i}}}|\;x\sigma_{i}x=\sigma_{i}^{z},\;z\in\mathbb{Z}\\}.$ (50)

By previous results, the subgroup $H_{\sigma_{i}}$ acts on the cycles in
$\sigma_{i}$ through the permutation representation
$H_{\sigma_{i}}/K_{\sigma_{i}}$ (Lemma 2.15).

######  Theorem 3.8.

The homomorphisms $\varphi,\psi$ are conjugate if and only if:

  1. (i)

$\mu_{0},\mu_{0}$ are conjugate in $S_{n}$.

  2. (ii)

$\pi_{i},\pi_{i}^{\prime}$ are conjugate in $S_{n}$ and the elements also have
the same cycle type under $H_{\sigma_{i}}/K_{H_{\sigma_{i}}}$, $1\leq i\leq
l$.

###### Proof.

The direct product decomposition of $\varphi(s),\psi(s)$ due to invariance of
cycle type and Lemma 2.18. ∎

We finally note that the relationship between element-conjugacy and
homomorphism conjugacy for the dihedral case is similar to the abelian case
with two generators:

######  Corollary 3.9.

The homomorphisms $\varphi,\psi$ are conjugate if and only if they are
element-conjugate on restriction to $S_{\mathrm{Fix}(\sigma)}$ and
$H_{\sigma_{i}}$, for $1\leq i\leq l$.

###### Proof.

If the order of $r$ is odd, the result follows from Corollary 2.19. If $d$ is
even, we proceed by contradiction to show that if $\pi_{i},\pi_{i}^{\prime}$
have distinct cycle types under $H_{\sigma_{i}}/K_{H_{\sigma_{i}}}$, on
restriction to $H_{\sigma_{i}}$ the elements $\varphi(rs),\psi(rs)$ are not
conjugate in $S_{n}$. ∎

## References

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  * M.Larsen [1996] M.Larsen.  On the conjugacy of element-conjugate homomorphisms ii.  _Quart. J. Math. Oxford_ , 47:73–85, 1996.
  * M.Weidner [2020] M.Weidner.  Pseudocharacters of homomorphisms into classical groups.  _Transformation Groups_ , 25:1345–1370, 2020.