If there is a simple group of order less than 36, then it must have prime order I want to show that if there is a simple group of order less than 36, then it must have prime order. 
Is there a quick way to show this or do I have to go through each order 1 though 36 showing that if it isn't prime, then it can't be simple?
Any help would be greatly appreciated. 
 A: You can do better than checking every single composite order up to $36$. We will make use of the following: It is a fact that if $G$ is a $p$-group, then $G$ is simple if and only if $\vert G \vert = p$. For a prime $p$ let $n_p$ be the number of $p$-Sylow subgroups of $G$. The Sylow theorems tell us that $n_p \equiv 1 \pmod p$ and $n_p$ divides $\vert G \vert$. Also, if there is a unique $p$-Sylow subgroup $P$ of $G$, then $P$ is normal. So if $\vert G \vert$ is not prime, then showing that there is a unique $p$-Sylow subgroup for some prime $p$ is enough to show that $G$ is not simple.
Let $G$ be a group with $\vert G \vert \leq 36$ composite. Then the largest prime dividing $\vert G \vert$ is at most $17$.
Suppose a prime $p \geq 7$ divides $\vert G \vert$. Then $\vert G \vert / p < 6$, hence $\vert G \vert$ is one of $2p, 2^2p, 3p$. Now $n_p \equiv 1 \mod p$ and $n_p$ divides $\vert G \vert$, so we must have $n_p = 1$.
We may thus assume the only prime divisors of $\vert G \vert$ belong to $\{2,3,5\}$. If $5$ divides $\vert G \vert$, then $\vert G \vert$ is one of $2\cdot 5, 2^2 \cdot 5, 2 \cdot 3 \cdot 5, 3 \cdot 5, 5^2$. Now we may exclude the case $\vert G \vert = 5^2$. The above style argument shows that there is a unique $5$-Sylow subgroup in the other cases except $\vert G \vert = 2 \cdot 3 \cdot 5$, in which case you can use a counting argument to show that one of the $p$-Sylow subgroups must be normal.
Excluding $p$-groups, we are thus left to the cases $\vert G \vert = 2 \cdot 3, 2^2 \cdot 3, 2^3 \cdot 3, 2 \cdot 3^2$. I will leave you to check these cases.
