Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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


Applying this to the above, we get


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


share|cite|improve this answer
+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. – Babak 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. – Babak 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. – user123454321 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. – Babak S. Jan 23 '13 at 3:01

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.

share|cite|improve this answer
Yes, but my question is to find a solution using $C_{G}(P)$. – user123454321 Jan 19 '13 at 3:47
And that's exactly why I thought knowing $n_{p}(G)$ is pretty much the answer. – user123454321 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. – user123454321 Jan 19 '13 at 3:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.