elementary proof on cyclic group

Question

Let $\xi,\gamma \in S_n$

Let $$\xi=(a_1 a_2 ... a_k)(b_1 b_2... b_l)...(g_1 g_2... g_q)$$ (disjunct cycle).

Prove that $$\gamma\xi\gamma^{-1}=\big((\gamma(a_1) \gamma(a_2) ... \gamma(a_k)\big)\big(\gamma(b_1) \gamma(b_2)... \gamma(b_l)\big)...\big(\gamma(g_1) \gamma(g_2)... \gamma(g_q)\big)$$

This is just an elementary group theory exercise and the proof is probably easier then I expect it to be. But I still don't know how to go for it.

My gut tells me to apply $\gamma^{-1}$ to the whole thing but I don't know if I'm allowed to do this or how it works. :/

Answer 1

You can do "the whole thing", but you are better considering what happens to a single element $\gamma(a_1)$ of the underlying set when you apply successively $\gamma^{-1}, \xi, \gamma$

Answer 2

When you want to write down the decomposition of a permutation $\sigma$ into disjoint cycles, you start with any element $a$, write $\sigma(a)$, then $\sigma(\sigma(a))$ and so on until you get back to $a$. At that point you close the cycle and restart with any element that didn't appear in the previous step.

In this case, the hint is to start from $\gamma(a_1)$. Then you know that $\xi(a_1)=a_2$, so $$ \gamma\xi\gamma^{-1}(\gamma(a_1))= \gamma\xi(a_1)=\gamma(a_2). $$ Then you go on in the same way: note that $\gamma(a_1),\gamma(a_2),\dots,\gamma(a_k)$ are all distinct and that $$ \gamma\xi\gamma^{-1}(\gamma(a_k))= \gamma\xi(a_k)=\gamma(a_1) $$ so the first cycle ends at $\gamma(a_k)$ and is indeed $$ \bigl(\gamma(a_1)\,\gamma(a_2)\,\dots\,\gamma(a_k)\bigr). $$

Then start from $\gamma(b_1)$ (which surely didn't appear before) and go on in the same way.

Answer 3

The simple idea in the proof is following: consider bijective functions $\sigma,\tau$ on any set, for example permutations you considered. The idea is: $$\mbox{if $\sigma$ takes $i$ to $j$ then $\tau\sigma\tau^{-1}$ takes $\tau(i)$ to $\tau(j)$}, $$i.e. $$ \sigma(i)=j \Longrightarrow \tau\sigma\tau^{-1}(\tau(i))=\tau(j).$$ Prove the last very easy implication; then you can immediately see your statement in question.