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

• February 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 $|G| = mp$ where $p$ is a prime and $1 < m < p$. Prove that $G$ is not simple.
• February 28th 2011, 05:17 PM
Tinyboss
How many Sylow p-subgroups can G have?
• February 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?
• February 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.
• February 28th 2011, 07:01 PM
JJMC89
Since $p|mp$, there is a a subgroup $H$ of order $p$. A (sub)group of order $p$ is cyclic, thus $H$ is Abelian thus normal. Since we have a normal subgroup $H\ne\{e\}$ and $H\ne G$, $G$ is not simple.

Is this argument correct?
• February 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.
• March 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.
• March 1st 2011, 10:26 PM
Drexel28
Quote:

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

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