Is it always true that $(a,b,c)(a,b,c) = (a,c,b)$? I noticed that $(1,2,3)(1,2,3) = (1,3,2)$, and I also noticed that $(1,4,3)(1,4,3) = (1,3,4)$. Now, my question is whether or not it is true that for any permutation $(a,b,c)^2 = (a,c,b)$?
 A: This is true for 3 cycles because a 3 cycle generates a cyclic subgroup of order 3, which has only the 3 cycles $(a,b,c)$ and $(a,c,b)$, and given any nonidentity element $x$ of this subgroup, $x^2$ is the other nonidentity element. In particular, $(a,c,b)^2=(a,b,c)$.
A: Here's a purely computational answer. Since a permutation of a set is uniquely determined by its action on that set, let us consider how $(a,b,c)^2$ acts on $S$. Take $s \in S$. If $s$ is not one of $a,b,c$, then $(a,b,c)^2$ fixes $s$. Otherwise, we compute
$$a \mapsto b\mapsto c$$
$$b \mapsto c \mapsto a$$
$$c \mapsto a \mapsto b$$
so $(a,b,c)^2 = (a,c,b)$.
A: I assume you mean "for any cycle of length three."
The answer is yes. Just follow what happens to any element when you apply $(a,b,c)$ to it twice, and compare that with what happens to the same element when you apply $(a,c,b)$ to it once. You'll have four cases to look at: if the element is $a$, if it's $b$, if it's $c$, and if it's something that's none of those.
A: Ofcourse this is true notice that you can also see it since we know that the order of a cycle is n, so you could see it that way aswell.
