Suppose that $a$ has order $h \pmod{p}$ and $a\overline{a} \equiv 1 \pmod{p}$. Show $\overline{a}$ also has order $h$.
I'm a little confused as to how to start proving this -- I know that if $a$ has order $h \pmod{p}$, then the positive integers $k$ such that $a^k \equiv 1 \pmod{p}$ are precisely those for which $h|k$. This would lead me to conclude that if I can prove $a|\overline{a}$, then I'm done. But is this the correct approach? Thanks.
We know that $a^h \equiv 1$. Suppose that $\overline a$ does not have order $h$, and consider the product $(a \overline a)^h$.
On the one hand, this is $1^h \equiv 1$. On the other hand, this is $a^h \overline a ^h \equiv 1 \overline a^h$. So we conclude that $\overline a^h \equiv 1$. This doesn't necessarily mean that $\overline a$ has order $h$ yet - just that $\overline a$ has order dividing $h$.
So now we need to show that the order of $\overline a$ is not less than $h$. To do this, we use the same idea. Suppose the order of $\overline a$ is $d < h$, a divisor of $h$. Then $ 1 \equiv (a \overline a)^d \equiv a^d$. This is false, as the order of $a$ is $h$, and therefore not $d$. So the order of $\overline a$ is $h$ by contradiction. $\diamondsuit$
This proof works in generality, showing that in any group, the order of an element is the same as the order of its inverse.