I've been given the following question in the context of group actions through conjugation but I'm having difficulty understanding what is being asked

Let $\tau$ be any permutation in $S_m$.

Let $\sigma$ be a cycle $\sigma = (a_1a_2...a_n)$ in $S_m$. Show that $\tau\sigma\tau^{-1}$ takes $\tau(a_1) \rightarrow \tau (a_2)$, $\tau(a_2) \rightarrow \tau (a_3)$, $...$ ,$\tau(a_n) \rightarrow \tau (a_1)$. Hence $\tau\sigma\tau^{-1}=(\tau(a_1)\tau(a_2)...\tau(a_n))$.

I do not quite understand what $\tau(a_1)$ means. To me it seems that $\tau(a_1) = \tau$ since $(a_1)$ is a permutation of 1 item. I'm guessing $ \tau(a_1)$ can be thought of as a function? But I am not quite sure how to intepret this.

Any help would be appreciated.

  • 2
    $\begingroup$ $\tau$ is some permutation. $\tau (a_1)$ would mean the element to which $a_1$ is mapped to by $\tau$. $\endgroup$
    – vnd
    Oct 18, 2015 at 12:49
  • 3
    $\begingroup$ In this case, these are "function application parentheses". $\tau(a_1)$ is just the value of "$\tau$ evaluated at $a_1$". $\endgroup$ Oct 18, 2015 at 12:49

1 Answer 1


Regarding the proof (vs.the notation)

Note $A=\{\tau(a_1), \dots ,\tau(a_n)\}$ and $\mathbb N_m=\{1, \dots, m\}$.

For $x \in \mathbb N_m \setminus A$, you have $\tau^{-1}(x) \notin \{a_1, \dots ,a_n\}$. Hence $\sigma \tau^{-1}(x)=\tau^{-1}(x)$ and $x=\tau \sigma \tau^{-1}(x)=(\tau(a_1)\tau(a_2)...\tau(a_n))(x)$ as $x \notin A$.

While for $x=\tau(a_i)$ with $1 \le i \le n$: $$\tau \sigma \tau^{-1}(x)=\tau\sigma(a_i)=\tau(a_{i+1})=(\tau(a_1) \ \dots \ \tau(a_n))(\tau(a_i))$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.