# Thread: No simple groups of order 48

1. ## No simple groups of order 48

Prove there are no simple groups of order 48
Ok, here's what i have so far.
$48=2^4*3$
$n_{2} = 1$ mod $2$ & $n_{2} | 3$
$n_{2} = 1$ or $3$
$n_{3} = 1$ mod $3$ & $n_{3} | 16$
$n_{3} = 1, 4,$ or $16$

So we must eliminate these cases to get $n_{2} = 1$ or $n_{3} = 1$:
$(n_{2}, n_{3})=(3,4)$
$(n_{2}, n_{3})=(3,16)$

How do I eliminate these cases?

2. Originally Posted by frankdent1
Prove there are no simple groups of order 48
Ok, here's what i have so far.
$48=2^4*3$
$n_{2} = 1$ mod $2$ & $n_{2} | 3$
$n_{2} = 1$ or $3$
$n_{3} = 1$ mod $3$ & $n_{3} | 16$
$n_{3} = 1, 4,$ or $16$

So we must eliminate these cases to get $n_{2} = 1$ or $n_{3} = 1$:
$(n_{2}, n_{3})=(3,4)$
$(n_{2}, n_{3})=(3,16)$

How do I eliminate these cases?
This problem is a little involved. Consider the Sylow 2-subgroups we know that there are 1 or 3 such Sylow subgroups. If there was only one Sylow 2-subgroup then there is nothing to prove because it is invariant under conjugation and so it is normal. Thus, we will assume there are 3 Sylow 2-subgroups. Let $H,K$ be two of these three Sylow subgroups. Notice that $|H|,|K| = 16$ and $|HK||H\cap K| = |H||K|$. Since $HK\subseteq G \implies |HK| \leq 48$, therefore $|H\cap K|$ cannot be less than or equal to $4$. This forces $|H\cap K|=8$ by applying Lagrange's theorem. Since $H\cap K$ is of index $2$ in $H$ and $K$ it means $H\cap K \triangleleft H, H\cap K\triangleleft K$. Therefore, the normalizer, $N(H\cap K)$ must contain both $H$ and $K$. By Lagrange's theorem it means $|N(H\cap K)|>16$ is a multiple of $16$ and a divisor of $|G|=48$. The only possibility is $N(H\cap K) = G \implies H\cap K\triangleleft G$ and so $G$ is not simple.