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.

Possible duplicate of [Showing groups of order $p^{k}(p+1)$ are not simple, p prime](http://math.stackexchange.com/questions/24855/showing-groups-of-order-pkp1-are-not-simple-p-prime) — 2017-01-10
Yeah , but can it be done in a more simpler way.actually I did not get that. — 2017-01-10
Actually , I will be highly grateful if anyone writes the complete answer for me.I am facing problems from underatanding those comments. — 2017-01-10
The detailed answer to the linked question has an argument for $k>1$, saying that the case $k=1$ needs a different one. — 2017-01-10

user id:58401 58k · Accepted Answer

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.

@mathislove, [Normalizer and centralizer](https://en.wikipedia.org/wiki/Centralizer_and_normalizer). — 2017-01-10
@mathislove, you need a basic understanding of [Sylow theorem's](https://en.wikipedia.org/wiki/Sylow_theorems) to go through this kind of solution. — 2017-01-10
The points of normaliser is in P but how are they exactly equal ? Same for cebtralizer — 2017-01-10

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