A problem about group of order $p(p+1)$.

Question

Let $G$ be a group of order $p(p+1)$ where $p$ is an odd prime and $n_{p}(G) = |\text{Syl}_{p}(G)| > 1$. The problem is to count the number of elements of $G$ that do not have order $p$.

The question I have is not just how to solve it but to understand a solution that is different from what I have thought. For me, it was easy to see that the number of elements that has order $p$ is $(p-1)n_{p}(G)$. Let $x$ be an element of order $p$ and write $P := \langle x \rangle$. Using the Sylow conjugacy theorem and the fact that $n_{p}(G) \equiv 1$ mod $p$, it is not hard to see that $n_{p}(G) = p + 1$ (since $n_{p}(G) = |G|/|N_{G}(P)|$), so the answer to the problem should be $p + 1$.

The solution that I was looking at, however, said to "consider $C_{G}(P)$ and $N_{G}(P)$." And it said "it was clear that $n_{p}(G) = p + 1$" and argued that "$|C_{G}(P)| = |N_{G}(P)|$." I cannot find the solution. A person I know just showed me a couple of hours ago. Can anyone recover this solution? I apologize if I misremembered anything. But the main points are:

1. Is $n_{p}(G) = p + 1$ that clear? I am not asking if more mathematically-matured people think that this is easy to see. I think that this is pretty much the answer to the question, and it is not right to say an answer to a problem is just "clear" if one wants to talk about how to solve the problem.

2. How would one use the centralizer $C_{G}(P)$ to solve this problem?

Answer 1

Well, for sure we have (from Sylow theorems) that

$$n_p(G)=[G:N_G(P)]=\frac{|G|}{|N_G(P)|}=p+1\Longleftrightarrow |N_G(P)|=p\Longrightarrow N_G(P)\in Syl_P(G)$$

Now, for any finite subgroup $\,H\leq G\,$ in any group, it is true that

$$\left|N_G(H)/C_G(H)\right|\,\,\mid\,\,\left|\operatorname{Aut}(H)\right|$$

Applying this to the above, we get

$$\frac{p}{|C_G(P)|}=\left|N_G(P)/C_G(P)\right|\,\,\mid\,\,(p-1)$$

But $\,\forall\,P\in Syl_p(G)\,\,,\,\,P\,$ is abelian , so $\,P\leq C_G(P)\,$ , which forces, by the above,

$$|C_G(P)|=p=|N_G(P)|\,$$

Answer 2

Your answer is correct. Since by assumption $n_p(G) > 1$, Sylow's third theorem forces $n_p(G) = p + 1$. Then you have $p + 1$ groups of order $p$ which intersect only in ${e}$, giving $(p+1)(p-1)$ elements of order $p$. Then the elements which don't have order $p$ necessarily do not belong to one of these groups, and so there are $p(p+1) - (p+1)(p-1) = p+1$ of these. No idea why one should consider centralisers or normalisers.