Let C be a group with |C| = 44. Prove that C must contain an element of order 2. I got stuck on this for a bit, then came across Cauchy's theorem:
Let G be a finite group and let p be a prime that divides the order of G.  Then G has an element of order p.
G is obviously finite, and 2 is a prime that divides the group.  Is this all I need to prove it?  I feel that was too easy and am probably missing something.   Mind you, we have yet to get to that chapter so maybe "technically" I can't use it, but is it still valid to do so?
Thanks,
Jeff
 A: Hint:
For an elementary proof,  just factor $44$ and note  it has only $6$ divisors. Then  Lagrange's theorem tells you the order of an element is one of these divisors. Examine each case, observe that if there's no element of order $2$, there's no element of order an even number either, so all elements $\ne e$ have order $11$. Is it compatible with $|C|=44$?
A: Cauchy's theorem for $p=2$ has a simple elementary proof.

Let $G$ be a finite group of even order. Then $G$ has an element of order $2$.

The orbits of the map $x \mapsto x^{-1}$ partition $G$ into the subsets $\mathcal O(x)=\{x, x^{-1}\}$. These subsets have cardinality $1$ or $2$. In fact, $\mathcal O(x)$ has one cardinality $1$ iff $x^2=1$ iff $x=e$ or $x$ has order $2$.
Let $m_k$ be the number of such subsets that have cardinality $k$. If $G$ has order $n$, then $n=m_1+2m_2$. Since $n$ is even, so is $m_1$. Now, $m_1\ge1$ because of $|\mathcal O(e)|=1$. Therefore, $m_1 \ge 2$. In other words, there is an element of order $2$.
