Prove that in any group, an element and its inverse have the same order.
The Attempt at a Solution
I know that I need to show |g|=|g^-1|. This is how I approached it but I'm not sure I'm right, it seems like I'm missing something.
Let |g|=m and |g^-1|=n for g, g^-1 elements of G. Then (g)^m=e and (g^-1)^n=e. Therefore, either (g)^m <= (g^-1)^n or (g)^m >= (g^-1)^n. If (g)^m <= (g^-1)^n then e<=e and m<=n. If (g)^m >= (g^-1)^n then e>=e and m >= n. Therefore m = n and |g|=|g^-1|.