# G is a group with |G|=mp where p is a prime and 1<m<p. Prove that G is not simple.

• Feb 28th 2011, 02:30 PM
JJMC89
G is a group with |G|=mp where p is a prime and 1<m<p. Prove that G is not simple.
Suppose that G is a group with $\displaystyle |G| = mp$ where $\displaystyle p$ is a prime and $\displaystyle 1 < m < p$. Prove that $\displaystyle G$ is not simple.
• Feb 28th 2011, 05:17 PM
Tinyboss
How many Sylow p-subgroups can G have?
• Feb 28th 2011, 05:44 PM
JJMC89
Quote:

Originally Posted by Tinyboss
How many Sylow p-subgroups can G have?

My class has not covered Sylow p-subgroups.
There is a Sylow p-subgroup of order p, right? Are there others or just one?
• Feb 28th 2011, 05:54 PM
Tinyboss
There are always Sylow p-subgroups. In this case, it follows immediately from Sylow's theorems that there is a unique one. Unique Sylow subgroups are normal.

But as you say, you haven't covered that yet, so I'm sure you're expected to make a more direct argument.
• Feb 28th 2011, 07:01 PM
JJMC89
Since $\displaystyle p|mp$, there is a a subgroup $\displaystyle H$ of order $\displaystyle p$. A (sub)group of order $\displaystyle p$ is cyclic, thus $\displaystyle H$ is Abelian thus normal. Since we have a normal subgroup $\displaystyle H\ne\{e\}$ and $\displaystyle H\ne G$, $\displaystyle G$ is not simple.

Is this argument correct?
• Feb 28th 2011, 07:04 PM
Tinyboss
Abelian subgroups need not be normal. But I think you're on the right track considering the cyclic subgroup generated by an order-p element.
• Mar 1st 2011, 11:32 AM
JJMC89
Quote:

Originally Posted by Tinyboss
Abelian subgroups need not be normal. But I think you're on the right track considering the cyclic subgroup generated by an order-p element.

Right. The subgroups of an Abelian subgroup H are normal in H but necessarily normal in the group G.

I am stuck now.
• Mar 1st 2011, 10:26 PM
Drexel28
Quote:

Originally Posted by JJMC89
Suppose that G is a group with $\displaystyle |G| = mp$ where $\displaystyle p$ is a prime and $\displaystyle 1 < m < p$. Prove that $\displaystyle G$ is not simple.

Let $\displaystyle H\leqslant G$ be such that $\displaystyle |H|=p$. It is trivial that there is a homomorphism $\displaystyle \phi:G\to \text{Sym}\left(G/H\right)$ by having $\displaystyle \phi_g(aH)=gaH$. Moreover, one can prove that $\displaystyle \ker\phi\subseteq H$. Now, since $\displaystyle p$ is prime and $\displaystyle \ker\phi\leqslant H$ we must have that $\displaystyle \ker\phi=\{e\}$ or $\displaystyle \ker\phi=H$. Suppose that $\displaystyle \ker\phi=\{e\}$ then $\displaystyle \text{im}(\phi)$ is a subgroup of $\displaystyle \text{Sym}\left(G/H\right)$ of order $\displaystyle mp$ and so $\displaystyle mp\mid m!$ but since $\displaystyle p$ is prime and $\displaystyle m<p$ this is impossible. Thus, $\displaystyle H=\ker\phi$ and so $\displaystyle \{e\}\triangleleft H\triangleleft G$ so that $\displaystyle G$ isn't simple.