Let $t_{k}$ be the number of elements of order $k$ in the group $G$. It known that if $|P|=p$ (where $P$ is Sylow $p$-subgroup of $G$), then $t_{2p}$ is a multiple of $t_{p}$. Now let $|P|=p^{2}$, the number of Sylow $p-$subgroups be $1$ and the number of cyclic subgroups of order $p$ be $p+1$. Also let the number of cyclic subgroups of order $2$ be $p(p+1)/2$. Is it true the number of cyclic subgroups of order $2p$ is a multiple of $p+1$? Or: Is it true $t_{2p}$ is a multiple of $t_{p}$? Thanks in advance.

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    $\begingroup$ You asked the same question on Math StackExchange, where I have given a counterexample. $\endgroup$ – Derek Holt Jul 23 '12 at 11:39
  • $\begingroup$ @Derek Holt: My question asked 7 hours ago and you have given a counterexample 2 hours ago on Math StackExchange. $\endgroup$ – Niki Jul 23 '12 at 11:44
  • $\begingroup$ It's usually better not to ask the same question in several places at the same time. This leads to duplication of effort, so it shows a lack of respect for the time of the people you are asking to help you. At the very least, you should post a link to the other question when you cross-post. $\endgroup$ – Douglas Zare Jul 23 '12 at 20:40

Not sure that this is a research level question: however.... . Under these hypotheses, $p$ is odd. An involution is an element of order $2$. If $t$ is an involution of $G,$ then $t$ must normalize an element of order $p$ (in fact, must centralize an element of order $p$). For otherwise $[G:C_{G}(t)]$ is divisible by $p^{2}.$ But $G$ is assumed only to contain $\frac{p(p+1)}{2}$ elements of order $2.$ But $G$ has a normal Sylow $p$-subgroup. Hence $C_{G}(t)$ contains $1$ or $p+1$ subgroups of order $p$ ( $1$ if $t$ does not centralize $P$, $p+1$ if $t$ centralizes $p$). The number of cyclic subgroups of order $2p$ maybe expressed as $\sum_{t} [G:C_{G}(t)] n_{p}(t)$, where $t$ runs over involutions of $G$ up to conjugacy, and $n_{p}(t)$ denotes the number of cyclic subgroups of $C_{G}(t)$ of order $p$. By assumption, $\sum_{t} [G:C_{G}(t)]$ is $\frac{p(p+1)}{2}$. This leaves uncounted $\sum_{u}p[G:C_{G}(u)]$ subgroups of order $2p,$ where $u$ runs over involutions of $C_{G}(P)$ up to $G$ conjugacy. I see no a priori reason why this last number should be divisible by $\frac{p+1}{2},$ which it would need to be for your question to have a positive answer.

  • $\begingroup$ Since Derek Holt has given a counterexample, this partial answer is redundant. $\endgroup$ – Geoff Robinson Jul 23 '12 at 12:11

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