α² is a cycle if and only if s is odd let $\alpha$ be a cycle of length $s$, say $\alpha = (a_1, a_2, \ldots, a_s)$
Prove $\alpha^2$ is a cycle if and only if $s$ is odd.
Let me start off by saying I am in my 5th week of Group Theory.  I often have trouble getting these problems started.  This is my first proof based course.  
I believe $\alpha^2 = (a_2, a_3, \ldots, a_1)$
Any tips on where to go from here would be great.  Also...if there are any tips for starting proofs like these in general, I could really use them! My teacher teaches as if a proofing class was a pre-req, which it was not. 
 A: HINT: Look at a couple of examples: $(1234)$ sends $1$ to $2$ and $2$ to $3$, so $(1234)^2$ sends $1$ to $3$. $(1234)$ sends $2$ to $3$ and $3$ to $4$, so $(1234)^2$ sends $2$ to $4$. $(1234)$ sends $3$ to $4$ and $4$ to $1$, so $(1234)^2$ sends $3$ to $1$. Finally, $(1234)$ sends $4$ to $1$ and $1$ to $2$, so $(1234)^2$ sends $4$ to $2$. Put all the pieces together, and you find that $(1234)^2=(13)(24)$, which is not a cycle. Do the same thing with $(12345)$, however, and you find that $(12345)^2=(13524)$, which is a cycle. Work that one through in detail to be sure, and then try to generalize these ideas to the cases $s$ even and $s$ odd.
A: Here is my somewhat crude illustration of what happens when you multiply a cycle with itself.
If the cycle has odd length, it just changes the order in which vertices are visited. The pentagon on the left is the graph of a $5$-cycle. The star on the right is the graph of that cycle multiplied with itself.

If the cycle has even length, things are different: it splits in two.

When you see this clearly in your mind, it should not be hard to write a formal proof.
