A group of order $12$ must have an element of order $2$ proof question. I have proven that there cannot be elements of only order $4,6,$ and $12$, so I just want to check for order $3$.
Suppose there are only subgroups of order $3$.
If $|G| = 12$ and $|a| = 3$, then there are $4$ cosets exactly.
If each $\langle a \rangle$ has exactly $3$ elements, then they must at least share the identity element.
If they only have the identity element in common, then the cosets created do not partition the entire set because not all elements of $G$ are included in this partition.  Any additional repeated elements in the cosets imply the same as well.
Does this make sense?
 A: If you know Sylow, there is a subgroup of order 4, which contains an element of order 2.
Your argument is good, you can also write it like so:
If you dont know Sylow.Take $x\in G$ $n(x)$ the order of $x$ divides 12 if it is even, then the order of $x^{n(x)/2}$ is 2. If the order of every element is odd it is 3. Each element generates a subgroup $G(x)$ of order 3, if $y$ is not in $G(x), G(x)\cap G(y)=1$, thus $|G|=1+2n$ contradiction
A: Very elementary proof that works for all groups of even order:
1) The number of fixed points of an involution of a finite set S has the same parity as S
2) Inversion is an involution in any group
3) The neutral element is a (trivial) fixed point of inversion
4) Therefore, in a group of even order inversion has at least one non-trivial fixed point
5) Any non-trivial fixed point of inversion is an element of order 2
A: You can use Cauchy theorem: for every prime $p \mid |G|$ there is a subgroup of order $p$. Because both $2,3$ divide $12$ there are two subgroups of order $2$ and $3$ respectively 
A: The number of elements of odd order is odd in any group as  every non-identity element can be paired with its inverse.
We conclude a group with an even number of elements must have at least one element with even order, and if $x$ has order $2k$ then $x^k$ has order $2$.
