If $g$ is a permutation in $S_n$, and $(a_1⋯a_k)$ is a $k$-cycle, show that $g(a_1⋯a_k)g^{−1}=(g(a_1)⋯g(a_k))$

Question

I am unsure how to go about solving this proof.

Thanks!

Answer 1

It's element tracing: two functions are equal iff they agree on every point of the domain : On the right hand side we have the function that sends the element $g(a_i)$ to $g(a_{i+1})$ for $1 \le i < k$ and $g(a_k)$ to $g(a_1)$ and all other points of $\{1,2,\ldots,n\}$ are fixed; this is by definition.

Now look at what the composition on the left does with the same elements. E.g. $g(a_i)$ is mapped to $a_i$ by $g^{-1}$ first, then the middle cycle sends it to $a_{i+1}$, and the final $g$ sends it to $g(a_{i+1})$. Similarly, $g(a_k)$ goes to $g(a_1)$. And if $i$ is any other element, not in the right hand cycle, then after the rightmost function we get $g^{-1}(i)$, which is not in the middle cycle ($g^{-1}(i)=a_j$ for some $j$ implies that $i=g(a_j)$, which we assumed is not the case) so is left alone by that cycle-map, and the leftmost $g$ then sends it back to $g(g^{-1}(i))=i$.

So on all $i$ in $\{1,2,\ldots,n\}$, the functions $g(a_1\,a_2 \cdots a_k) g^{-1}$ and $(g(a_1)\,g(a_2)\cdots g(a_k))$ have the same values so they are the same function.