The conjugate of a k-cycle by a transposition

Question

Is it true that in any symmetric group the conjugate of a k-cycle by a transposition is always a k-cycle?

Answer 1

The answer is yes, and in general given $\alpha, \beta \in S_n$, if $\alpha$ is the product of $k$ cycles of lengths $\ell_1,..., \ell_k$, then $\beta\alpha\beta^{-1}$ has the same cyclic structure as $\alpha$.

To prove this in the case where $\beta$ is a trasposition and $\alpha$ is a $t$-cycle, we distinguish $3$ cases:

• If $\alpha$ and $\beta$ are disjoint cycles, they commute, thus $\beta\alpha\beta^{-1}=\beta\beta^{-1}\alpha=\alpha$ is obviously a $t$-cycle.
• If $\alpha$ and $\beta$ share $1$ element, let $\alpha=(a_1,...,a_t)$ and $\beta=(a_k,b)$. Then, $\beta\alpha\beta^{-1}$ fixes $a_k$ and $$\beta\alpha\beta^{-1}=(a_1,...,a_{k-1},b,a_{k+1},...,a_t)$$ is indeed a $t$-cycle.
• If $\alpha$ and $\beta$ share $2$ elements, let $\alpha=(a_1,...,a_t)$ and $\beta=(a_m,a_n)$, where WLOG $m<n$. Then, $$\beta\alpha\beta^{-1}=(a_1,...,a_{m-1},a_n,a_{m+1},...,a_{n-1},a_m,a_{n+1},...,a_t)$$ is indeed a $t$-cycle.

To prove this in the general case, let $\alpha=\alpha_1\alpha_2\cdots \alpha_k$ where $\alpha_i$ is a cycle of length $\ell_i$. Since $$\beta\alpha\beta^{-1}=(\beta\alpha_1\beta^{-1})(\beta\alpha_2\beta^{-1})\cdots (\beta\alpha_k\beta^{-1}),$$ it suffices to prove that $\beta\alpha_i\beta^{-1}$ is a $\ell_i$-cycle for each $i=1,...\,,k.$ Since every permutation can be written as a product of traspositions, let $\beta=\beta_1\beta_2\cdots \beta_s,$ where each $\beta_j$ is a trasposition. Then, $$\beta\alpha_i\beta^{-1}=\beta_1\beta_2\cdots \beta_s\alpha_i \beta_s\cdots \beta_2\beta_i$$ is a $\ell_i$-cycle since we can inductively use the fact we just proved that the conjugate of a $t$-cycle by a trasposition is a $t$-cycle.

Answer 2

Yes! Indeed the conjugation of a $k$-cycle by anything is a $k$-cycle! (cf. this article)

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Hope this helps ^_^

Answer 3

Yes. In fact, it is frequently useful in computations to have in mind the following lemma, which specifies how $k$-cycles conjugate to $k$-cycles.

$$ \tau (a_1,\ldots,a_n) \tau^{-1} = (\tau(a_1),\ldots,\tau(a_n)) $$