Help proving a property of a group of order $4p$ I am having trouble with this very small proof and it is driving me crazy. I was hoping someone could give me some help:
If $G$ is a noncyclic group of order $4p$ for some odd prime $p \geq 5$, then $G$ contains an element of order $2$.
Thanks.
 A: This is to me a very strange formulation of the problem. Any group $G$ with $|G|$ even has an element of order $2$. No need for $G$ to be noncyclic or the other condition. This is a special case of Cauchy's Theorem, but for the case at hand a slightly simpler proof than (any one of) the standard proofs of Cauchy's Theorem is given as follows.
Consider the action of the group ${1,-1}$ on $G$ given by $ag=g^a$. Notice that an element $g\in G$ has orbit size equal to $1$ iff $g^2=e$. Since the orbits partition $G$ and each orbit has size either $1$ or $2$, and $|G|$ being even, if there is one orbit of size $1$ there has to be another one. Since the orbit of $e$ has size $1$ it follows there is some $g\in G$ whose orbit has size is $1$, thus $g^2=e$ and an element of order $2$ is found. 
A: Suppose the finite group $G$ has an even number of elements. (Note: clearly $4p$ is even.)
Imagine a Cayley table (a group multiplication table). 
Suppose for the sake of contradiction that  no element has order $2$. 
Then there is only one $e$ along the NW/SE diagonal of the table; namely, for $e*e = e$.
But we know that $xy = e$ implies $yx = e$ as well, so every other $e$ in the table comes in a pair symmetric around the aforementioned diagonal.
Therefore, the number of $e$'s in the table is odd: the one in the diagonal and the rest in pairs.
But the number of $e$'s in the table is also even: one in each row, and $|G|$ is even.
Contradiction.
Thus, we were wrong in our supposition, and there must be an element of order $2$.
A: If the group has any elements whose order is a multiple of 2, then you're done. So the only way you could run into trouble is if every nontrivial element has order p. Now apply the theorem you mention in comments.
To my mind this is a slightly odd way of ending the argument. If every nontrivial element has order p, then you have 4p-1 nontrivial elements grouped into subgroups each with p-1 nontrivial elements and no overlaps.  But then $(p-1) \mid (4p-1)$, so $(p-1) \mid 3$, which is impossible because p is odd.
A: Suppose that there are no elements of order $2$. Then, there are no elements of order $4$ either, because a cyclic group of order $4$ has one element of order $2$. Since $G$ is noncyclic, there's also no one element of order $4p$. So, we are left with all the nontrivial elements having order $p$. But then $4p=k(p-1)+1$, for some integer $k$, whence:
\begin{alignat}{1}
k &= \frac{4p-1}{p-1} \\
&= \frac{3p}{p-1}+1 \\
\end{alignat}
which is not an integer whatever the odd prime $p$ is ($3p$ is odd, while $p-1$ is even): contradiction. (Seemingly, the condition $p\ge 5$ is unnecessarily tight, as the contradiction holds for $p=3$, too.)
