# Conjugate Permutations have Same Cycle Type

## Theorem

Let $n \ge 1$ be a natural number.

Let $G$ be a subgroup of the symmetric group on $n$ letters $S_n$.

Let $\sigma, \rho \in G$.

Then $\sigma$ and $\rho$ are conjugate if and only if they have the same cycle type.

## Proof

Let $\sigma \in G$ have cycle type $\tuple {k_1, k_2, \ldots, k_n}$.

From Existence and Uniqueness of Cycle Decomposition, $\sigma$ can be expressed uniquely as the product of disjoint cycles:

- $\sigma = \alpha_1 \alpha_2 \dotsm \alpha_l$

where $\alpha_i$ is a $k_i$-cycle.

Let $\tau \in G$ such that $\rho = \tau \sigma \tau^{-1}$.

Then:

\(\displaystyle \tau \sigma \tau^{-1}\) | \(=\) | \(\displaystyle \tau \alpha_1 \alpha_2 \dotsm \alpha_l \tau^{-1}\) | |||||||||||

\(\text {(1)}: \quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle \tau \alpha_1 \tau^{-1} \tau \alpha_2 \tau^{-1} \dotsm \tau \alpha_l \tau^{-1}\) | Product of Conjugates equals Conjugate of Products | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \alpha_{\map \tau 1} \alpha_{\map \tau 2} \dotsm \alpha_{\map \tau l}\) | Cycle Decomposition of Conjugate |

We have that for $i, j \in \set {1, 2, \ldots, l}$, $\alpha_i$ and $\alpha_j$ are disjoint.

We also have that $\tau$ is a bijection, and so an injection.

It follows that $\alpha_{\map \tau i}$ and $\alpha_{\map \tau j}$ are also disjoint for all $i, j \in \set {1, 2, \ldots, l}$.

Thus:

where:

- $\tau \alpha_i \tau^{-1}$ is a $k_i$-cycle.

Thus $\rho$ has the same cycle type as $\sigma$.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 80 \alpha$