G is not simple.

Question

Let $G$ be a group of order $p\cdot (p+1)$ , where $p$ is an odd prime.

How can I show , that $G$ is not simple ?

I am stuck here.I tried with all the tools I have but I failed . I also tried sylow but could not get a way.please anyone help me.Thanks in advance.

Answer 1

Suppose $G$ simple.

Hint 1

The number of $p$-Sylow subgroups is $p+1$, so that if $P$ is one of these subgroups, we have $N_{G}(P) = P = C_{G}(P)$.

Hint 2

There are $(p-1)(p+1) = p^{2} -1$ elements of order $p$.

Hint 3

There are $p+1$ elements left. Let $\Omega$ be the set of the non-identity elements among them, so $\Omega$ has $p$ elements.

Hint 4

$P$ acts by conjugation on $\Omega$. Since $C_{G}(P) = P$, $P$ acts transitively on $\Omega$ by conjugation.

Hint 5

All the elements of $\Omega$ have the same order, so $p+1$ is prime.

Hint 6

$p$ is odd.