Is my proof of G being a finite set of even order $\iff\exists g\in G$ with order 2 okay? Is my proof of G being a finite set of even order $\iff\exists g\in G$ with order 2 okay? (with $g\ne1$ obviously)
There are loads of duplicates of this question but none have "proved" it like I have, as such I'm panicky. I opted for a proof by contradiction. 
I prove it as follows:
(1 denotes the identity, I'm using multiplicative notation)
An element of order 2 means $g^2=1$ which $\iff$ $g^{-1}=g$. 
Let us assume that no such element exists, then $\forall g\in G, g\ne 1$ we have $g\ne g^{-1}$ (else it would be of order 2), however $(g^{-1})^{-1}=g$, so $|g|=|g^{-1}|$ ($|g|$ is the order of g)
Because $g\ne g^{-1}$ g and its inverse are distinct. Thus we can write our group as:
$1,g_1,g_1^{-1},\dots,g_n,g_n^{-1}$ 
We know there's an $n$ because the group is stated to be finite.
counting these distinct elements we see that there is the identity, and n pairs, meaning:
$|G|=2n+1$ - which is the very definition of an odd number.
So I have shown if a group has no element of order 2 than it must be odd, so it follows that to have an element of order 2 it must be of even order.
I am not saying all even groups have an element of order 2, I am saying that to have an element of order 2 it must be even.
However the simple tranposition that takes 1 to 2 and 2 to 1 is of order 2. Which is present in every symmetry group for n$\ge$2, and they all have even order ($|S_n|=n!$) but this feels a bit wooley. 
Have I proved that all finite groups of even order have an element not equal to the identity that has order 2? I think so!
side note
I'm a bit uncomfortable with the n I pulled from the group being finite because I then showed the order to be 2n+1, is that "okay" because it seems a bit circular. 
 A: You have shown
$$\text{no element of order two}\implies\text{group has odd order}.$$
and conclude that "to have an element of even order the group must have even order." But you've actually shown something stronger than this: the contrapositive of the above implication is
$$\text{group has even order}\implies\text{there is an element of order two}.$$
Not only is it necessary for the group to have even order for there to be an element of order two; it is also sufficient for the group to have even order. (Note the contrapositive of any implication is logically equivalent to the original implication ... not so with converses.)
Also note that the order of an element is the minimal positive exponent needed to get it to make the identity element. This means the identity element itself has order $1$, not order $2$. Thus, an element has order two if and only if it satisfies both $g^2=e$ and $g\ne e$. So if you state "$g$ has order two," it is automatic that $g\ne e$; you don't have to tack that condition on separately because it is built-in.
