An element of a group has the same order as its inverse 
If $a$ is a group element, prove that $a$ and $a^{-1}$ have the same order.

I tried doing this by contradiction.
Assume $|a|\neq|a^{-1}|$.
Let $a^n=e$ for some $n\in \mathbb{Z}$ and $(a^{-1})^m=e$ for some $m\in \mathbb{Z}$, and we can assume that $m < n$.
Then $e= e*e = (a^n)((a^{-1})^m) = a^{n-m}$. However, $a^{n-m}=e$ implies that $n$ is not the order of $a$, which is a contradiction and $n=m$.
But I realized this doesn’t satisfy the condition if $a$ has infinite order. How do I prove that piece?
 A: Let $a^n$ be $e$, then $e=(aa^{-1})^n=a^n(a^{-1})^n=e(a^{-1})^n=(a^{-1})^n$.
Let $(a^{-1})^n=e$, then $e=(aa^{-1})^n=a^n(a^{-1})^n=a^ne=a^n$.
So, $a^n=e \iff (a^{-1})^n=e$.
A: Suppose that $a$ has infinite order. We show that $a^{-1}$ cannot have finite order. Suppose to the contrary that $(a^{-1})^m=e$ for some positive integer $m$. We have by repeated application of associativity that
$$a^m (a^{-1})^m=e.$$
It follows that $a^m=e$.
A: Let's say
a is an element and n is its order ,then
$$a^n=e$$
Repeatedly multiplication by $a^{-1}$ n times
$$(a^{-1})^{n}•(a)^{n}=e•(a^{-1})^n$$
$$(a^{-1}•a)^{n}=(a^{-1})^n$$
$$e^n=(a^{-1})^n=e$$
Hence "a" 's inverse is also having order of n.
A: It seems a more straightforward solution exists?
If $g$ has infinite order then so does $g^{-1}$ since otherwise, for some $m\in\mathbb{Z}^+$, we have $(g^{-1})^m=e=(g^m)^{-1}$, which implies $g^m=e$ since the only element whose inverse is the identity is the identity. This contradicts that $g$ has infinite order, so $g^{-1}$ must have infinite order.
If $g$ has finite order $n$, then by existence of inverses in a group $$g^n=e\iff$$ $$g^n \cdot (g^{-1})^n=e\cdot(g^{-1})^n\iff$$ $$g^n\cdot g^{-n}=(g^{-1})^n\iff$$ $$ e = (g^{-1})^n$$ This implies $|g^{-1}|\leq n$.
If $|g^{-1}|<n$, say $m$, then $(g^{-1})^m=e=(g^m)^{-1}\implies g^m=e$, which contradicts that $|g|=n>m$. So $|g^{-1}|=n$ if $|g|=n$.
