I'm stuck on a review question for my Abstract Alg. exam. The question is:
Suppose $A$ and $B$ are permutations such that when written is cycle form, they have no symbols in common. Prove that if $A^k$ = $B^k$, then both $A^k$ and $B^k$ equal $e$ (the identity).
I have no idea how to get started on this problem. If $A$ and $B$ have no symbols in common in cycle form, then does that mean they are made of different disjoint cycles? Any tips on how to do this problem would be greatly appreciated. Thanks.
We can prove that if $A^k = B^j$, then $A^k = B^j = e$, for any integers $k,j$. Let $A$ move the symbols $a_1, a_2, \ldots, a_n$ and $B$ move the symbols $b_1, b_2, \ldots, b_m$. Let $H$ and $K$ be the subgroups generated by $A$ and $B$ respectively. Then $H$ is a subgroup of the symmetric group on $a_1, a_2, \ldots, a_n$ and $K$ is a subgroup of the symmetric group on $b_1, b_2, \ldots, b_m$. Since $A$ and $B$ have no symbols in common, it follows that $H \cap K = (e)$. Thus $A^k = B^j$ implies $A^k = B^j \in H \cap K$ and hence is the identity.