TITLE: Is my proof correct? (group theory, Lagrange)
QUESTION [5 upvotes]: $G$ is a finite group.
Prove, that $\exists k\in G:k\neq 1_G, k^2=1_G \iff |G|$ is even.
My proof
1) $\Rightarrow$:
$\exists k\in G: k^2 = 1_G\Rightarrow \exists G'\leq G$ s.t. $ G'=\{1_G,k\}$, then $|G'|=2$
Using Lagrange Theorem:
$|G'|$ divides $|G|\Rightarrow |G|$ is even.
2) $\Leftarrow$:
$|G|=2n\Rightarrow |G|\geq 2$ and $|G|=2n\Rightarrow\exists G'\leq G$ with $|G'|=2\Rightarrow\{1_G=1_{G'}, k\}= G'$ and $k=k^{-1}\Rightarrow |k|=2$
$\blacksquare$
Could you please review my proof? I'm currently learning how to write proofs using as few words as possible, so could you also check if the syntax is correct?

REPLY [6 votes]: For the reverse implication I think you were trying to express this:-
If for all the elements $x\neq e$ in $G$, $x\neq x^{-1}$ then including identity total number of elements in $G$ is odd, which is not the case, so there must exist a non-identity element in $G$ which is its own inverse.