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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?

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2 Answers 2

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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,

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

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+1 Using $N/C$ lemma in a nice way. Could we say: If the group $G$ hasno normal subgroup of order $p$ then $n_p>1$ and since $n_p\mid p(p+1)$ and $n_p\equiv 1~~(\text{mod}~p)$ then $n_p=p+1$? Do you think I can add this answer as mine? Thanks Don. –  B. S. Jan 20 '13 at 3:29
    
Well, that covers the given info. I'm not sure if that qualifies as an answer but you could add a comment saying "the conditions of the problem exist when there's no normal Sylow $\,p-$subgroups", though perhaps this is so obvious that the OP isn't interested... –  DonAntonio Jan 20 '13 at 3:32
    
Thanks for your time. –  B. S. Jan 20 '13 at 3:34
    
@BabakSorouh What you wrote is exactly what I thought to be a solution to this problem, and again the solution that I saw thought that the argument for $n_{p} = p + 1$ was "clear". It went on arguing that $|C_{G}(P)| = |N_{G}(P)|$. All I wanted to know was pretty much this identity, not how to solve this problem. –  GYC Jan 23 '13 at 1:23
    
@GilYoungCheong: I got that point already throughout your problem. However, after that comment I saw the Don's answer neat and completing. Thanks for your consideration to my small comment. –  B. S. Jan 23 '13 at 3:01
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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.

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Yes, but my question is to find a solution using $C_{G}(P)$. –  GYC Jan 19 '13 at 3:47
    
And that's exactly why I thought knowing $n_{p}(G)$ is pretty much the answer. –  GYC Jan 19 '13 at 3:47
    
Well okay, then this is an answer to part 1 of your question. This solution is so simple that I don't know why you care about a solution with centralisers -- it's almost certainly more complicated. –  Joshua Ciappara Jan 19 '13 at 3:48
    
Hmm. I guess it "is" clear. Thanks for answering. –  GYC Jan 19 '13 at 3:50
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