# 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.
$\displaystyle 48=2^4*3$
$\displaystyle n_{2} = 1$ mod $\displaystyle 2$ & $\displaystyle n_{2} | 3$
$\displaystyle n_{2} = 1$ or $\displaystyle 3$
$\displaystyle n_{3} = 1$ mod $\displaystyle 3$ & $\displaystyle n_{3} | 16$
$\displaystyle n_{3} = 1, 4,$ or $\displaystyle 16$

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

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.
$\displaystyle 48=2^4*3$
$\displaystyle n_{2} = 1$ mod $\displaystyle 2$ & $\displaystyle n_{2} | 3$
$\displaystyle n_{2} = 1$ or $\displaystyle 3$
$\displaystyle n_{3} = 1$ mod $\displaystyle 3$ & $\displaystyle n_{3} | 16$
$\displaystyle n_{3} = 1, 4,$ or $\displaystyle 16$

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

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 $\displaystyle H,K$ be two of these three Sylow subgroups. Notice that $\displaystyle |H|,|K| = 16$ and $\displaystyle |HK||H\cap K| = |H||K|$. Since $\displaystyle HK\subseteq G \implies |HK| \leq 48$, therefore $\displaystyle |H\cap K|$ cannot be less than or equal to $\displaystyle 4$. This forces $\displaystyle |H\cap K|=8$ by applying Lagrange's theorem. Since $\displaystyle H\cap K$ is of index $\displaystyle 2$ in $\displaystyle H$ and $\displaystyle K$ it means $\displaystyle H\cap K \triangleleft H, H\cap K\triangleleft K$. Therefore, the normalizer, $\displaystyle N(H\cap K)$ must contain both $\displaystyle H$ and $\displaystyle K$. By Lagrange's theorem it means $\displaystyle |N(H\cap K)|>16$ is a multiple of $\displaystyle 16$ and a divisor of $\displaystyle |G|=48$. The only possibility is $\displaystyle N(H\cap K) = G \implies H\cap K\triangleleft G$ and so $\displaystyle G$ is not simple.