Property of odd ordered elements of a Group I'm (slowly) working my way through "Abstract Algebra" by Dummit and Foote. In the first set of exercises on group theory, the following question is posed:
"Let $G$ be a finite group and let $x$ be an element of $G$ of order $n$. Prove that if $n$ is odd, then $x = (x^2)^k$ for some $k$."
I believe I have completed the proof, but at no point have I relied (to my knowledge) on the fact that the group be finite. Note that no other exercises in this section are about finite groups, and it strikes me as odd that this be stipulated in the question. Leading me to believe I may have made a mistake or I have made an assumption that only applies to finite groups. Proof sketch below.
Any odd number $n$ can be decomposed into $2*j + 1$, for some $j$. Thus, taking $k = j + 1$:
$$\begin{align}
x &= 1x \\
  &= x^nx \\
  &= x ^{2j + 1}x\\
  &= x ^{2j + 2} \\
  &= x ^{2(j + 1)} \\
  &= (x ^{2})^{j+1}
\end{align}$$
Edit: To be clear, is my proof valid regardless as to whether the group is finite or not? If so, why do the author's specify finiteness?
 A: Your proof is correct. To answer your question, though, let's prove that this result (ignoring whether it is actually true or not), cannot be true only for finite groups. In other words, if it is true for finite groups, it must be true for infinite groups.
Let $H$ be a group of infinite order, and let $x \in H$ be an element of odd order $n$. Let $G = \langle x \rangle$. Then $x$ is an odd order element of a finite group (namely $G$). Thus, if the result holds for a finite group, it holds for an infinite group.
A: Your proof is correct, and very nice!
You asked why the question was phrased this way.  Indeed it could have been written differently:
"Let $G$ be a group and suppose $x$ is an element of order $n < \infty$.  Prove that if $n$ is odd, $\ldots$"
(If you assume that $G$ is finite, as the authors did, then you know automatically that every element has finite order, and you don't have to write $n< \infty$ explicitly.   Presumably the authors found their phrasing more elegant than what I've written, or stylistically preferable in some way.)
And you're right that you never used the fact that $G$ was finite.  However, the element $x$ is explicitly assumed to have finite order, which is equivalent to saying that the cyclic subgroup $\langle x \rangle$ has only finitely many elements.  So even though the group $G$ might just as well have been infinite, the subgroup $\langle x \rangle$ is definitely finite.
One way to think about this problem is that it's really a theorem about finite cyclic groups.  Of course a finite cyclic group might be a subgroup of a much larger (possibly infinite) group, but the theorem doesn't really have anything to do with the larger ambient group. In fact your proof shows this very clearly, since the only group elements that appear in your proof are powers of $x$.
A: Here is an explanation of what is going on.
If $n$ is odd, then the map $g \mapsto g^2$ is injective on the cyclic subgroup generated by $x$ (*). Since this subgroup is finite, the map is then surjective and so $x=(x^k)^2=(x^2)^k$.
(*) Indeed, if $(x^r)^2=(x^s)^2$ then $2r\equiv 2s \bmod n$. Since $n$ is odd, we can cancel $2$ and get $r\equiv s \bmod n$, which implies $x^r=x^ˆs$.
