Q:

Suppose the order of $g$ is even. What is the order of $g^2$?

Let $n=2k$ be the order of $g$. Then $g^{2k}=1,$ for some natural $k$. Then, $$(g^2)^k=g^{2k}=1,$$ so the order of $g^2$ divides the order of $g$.
Similarly, let $m$ be the order of $g^2$. Then, $(g^2)^m=1$. Then what?
Any pointers in proceeding would be appreciated.

A:

Generally speaking, there's a relation between the order of $g$ and the order of $g^i$:
$$\operatorname{order}(g^i)=\dfrac{\operatorname{order}(g)}{\gcd(\operatorname{order}(g),i)}.$$

A:

By definition, the order of $g^2$ is the smallest possible positive-integer value of $j$ such that $(g^2)^j = g^{2j} = 1$
We know setting $j=k$ satisfies this equation, so the order $m \leq k$. (In fact we can say $m|k$ -- we won't need that here, but think about why.)
On the other hand, if $m<k$, then $2m<2k$, and $g^{2m}=1$; what does this mean about the order of $g$, going back to the definition of order?

