Is my proof correct? (group theory, Lagrange) $G$ 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?
 A: 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. 
A: I think you're a little bit sloppy at the second part of the proof. Note that the premise is that $|G|$ is even (and $G$ is finite). The existence of $k$ is not part of the premise, but you seem to use it without motivation. 
Then of course formally ${1_G=1_{G'}, k}$ is probably not a member of $G'$, but possibly a subset of it.
The second part could for example be proven by noting that with $|G|\le2$ we have a non-identity element (generating a subgroup) that has either even or odd order. If it's even you can easily find the $k$ by taking the middle element. If it's odd you can use cosets to find a proper sub group $G'$ with $|G'|$ being even and repeat.
A: The implication in 2) $\Leftarrow$:  $|G|\geq2\Rightarrow\cdots$ is incorrect.
If e.g. $|G|=3>2$ then no $k\in G$ exists with $k\neq1_G$ and $k^2=1_G$.
The theorem of Cauchy can be applied here: 
If $p$ is a prime that divides the order of $G$ then $G$ contains an element of order $p$.
