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Let $G$ be a group with a non normal $p$-sylow subgroup $P$. Is there any information about the number of $p$-elements(an element with $p$-power order) of $G$?

Alexander Gruber
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Adeleh
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    In general, nothing more non-trivial than what's already in Sylow; one also needs an idea of the intersection of Sylow $p$'s; this is a rather delicate question, I'd say. But, you must first work out what Sylow tells you and ask more precise question. – knsam Aug 04 '13 at 09:48
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    I agree with kan. The question is too general. You need to ask something more specific if you want a helpful answer. (As a general rule, questions of the type "what can we say about ...?" should be made more specific.) – Derek Holt Aug 04 '13 at 10:18
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    The only situation where you can say something precise a priori is that of $p$ dividing exactly $|G|$. In this case the $p$-Sylows are cyclic of order $p$, and intersect each other trivially, so there are $n_p(p-1)$ elements of order $p$ in $G$. – Andrea Mori Aug 04 '13 at 11:00
  • I've asked some more specific versions of this question that might be interesting: http://math.stackexchange.com/questions/2355/is-there-a-group-with-exactly-92-elements-of-order-3 – Jack Schmidt Aug 04 '13 at 15:33
  • I have also asked a related question here – Mikko Korhonen Aug 04 '13 at 21:17
  • @m.k. good to know. That was actually a research direction of mine. Let me know if you are interested in my results. No one else was :-) – Jack Schmidt Aug 04 '13 at 21:23
  • @JackSchmidt: I'm interested! I spent a lot of time with this and other things related to solutions of $x^n = 1$.. are you going to post an answer or should I send you an email? – Mikko Korhonen Aug 04 '13 at 21:46

1 Answers1

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You can find out some general properties, but nothing precise really unless you have more information.

Let $G$ be a finite group. Denote by $a_p(G)$ the number of $p$-elements of $G$.

Frobenius' theorem states that when $n$ divides the order $G$, the number of solutions to $x^n = 1$ in $G$ is a multiple of $n$. Note that $a_p(G)$ is the number of solutions to $x^{p^\alpha} = 1$, where $p^\alpha$ is the largest power of $p$ dividing $|G|$. Hence Frobenius' theorem implies that $a_p(G)$ is a multiple of $p^\alpha$, say $a_p(G) = p^\alpha r$.

Now the number of elements of order $k$ in $G$ is a multiple of $\varphi(k)$, where $\varphi$ is the Euler totient function. Thus for any $1 \leq l \leq \alpha$, it follows that $p-1$ divides the number of elements of order $p^l$ in $G$. Hence $p-1$ divides $a_p(G) - 1$. In particular $p-1$ divides $r-1$, and thus

$$a_p(G) = p^{\alpha}(t(p-1) + 1)$$

for some integer $t \geq 0$.

Let $n_p(G)$ be the number of Sylow $p$-subgroups of $G$, so $n_p(G) \equiv 1 \mod{p}$ by Sylow's theorem. It is possible to deduce some small things about $a_p(G)$ if we know $n_p(G)$, see this question (SE) and this one (MO). For example, G. A. Miller proved that

  • If $n_p(G) = 1$, then $a_p(G) = p^\alpha$.
  • If $n_p(G) = p+1$, then $a_p(G) = p^{\alpha+1}$.
  • If $n_p(G) > p+1$, then $a_p(G) \geq p^\alpha(2p - 1)$.
Community
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Mikko Korhonen
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  • Your answer include some nice thing that I havn't known before. Thanks for your tips. Do you have any refrence for two last line? – Adeleh Aug 05 '13 at 06:55
  • @Adeleh: See this question. Specifically, Miller proves in pg 79-80 of "Theory and Applications of Finite Groups" (1916) that $a_p(G) > p^{\alpha+1}$ when $n_p(G) > p+1$. Now we can apply the fact that $a_p(G) = p^{\alpha}(t(p-1) + 1)$ to improve this to $a_p(G) \geq p^{\alpha}(2p-1)$. This improvement is from Miller's paper "Some Deductions From Frobenius's Theorem" (1942), it is near the end of the paper: link to paper. – Mikko Korhonen Aug 05 '13 at 07:54