Group with an even number of elements. If $G$ is a group such that $|G|=2n$. Prove that there's an odd number of elements of order 2, and then there's an element which is its own inverse, besides of the identity.
If we consider all the elements of $G$ that have order different than 2, we have two cases: 1) the elements with order greater than 2, and 2) elements with order 1.
For 1). Let $A=\{g\in G: g^k=e, k>2\}$, then if $g\in G$ then $g^{-1}\in G$ (should I prove that the inverse has the same order?), and also $e\not\in G$. Then there's an even number of elements in A, because $g\not = g^{-1}$. So $|A|=2m$ for some $m$.
For 2). The only element of order 1 is the identity, so $B=\{g\in G: g^k=e, k=1\}=\{e\}$ then $|B|=1$.
So $|A|+|B|=2m+1$ like we wanted, but if that happens, the second part of the problem doesn't make sense.
So doing this in a different way, instead of directly counting the elements of B, we consider the complement of A, so $|A^c|=2k$, that way $|G|=|A|+|A^c|=2n$ checks out right. But $A^c=\{g\in G: g^k=e, k\not>2\}=\{g\in G: g^k=e, k=1 or 2\}$, so because the identity is the only element of order 1, then there must be at least one element in $A^c$ of order 2, that means $g_o^2=g_o*g_o=e$ hence $g_o=g_o^{-1}$.
Is this right? What am I doing wrong in the first try?
I know this problem has been solved before in this site, but I really wanted to make it on my own, sorry if this is redundant.

Your first try is correct. By your proof, you know the number of elements of order 2 is odd, hence can not be zero. This means there exists at least one, say $g$, which is just the element you wanted in the second part.
