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Let $p$ be prime divisor of order of finite group $G$, and the number of cyclic subgroup of order $p$ be $p+1$. If $P$ is a Sylow $p$-subgroup of $G$, then $P$ is normal in $G$ and $|P|=p^{2}$($P$ is not cyclic). Also let the number of cyclic subgroup of order $2$ be $p(p+1)/2$. Is it true the number of cyclic subgroup of order $2p$ is a multiple of $p+1$?

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Where you say "...then P is normal in G and ...", you mean this is a given piece of data, right? I mean, because it doesn't follow from what you gave in the first line, as it seems to be implied... – DonAntonio Jul 22 '12 at 14:29
$A_4$ is a counterexample if you allow $p = 2$, although presumably you want to disallow this. – user29743 Jul 22 '12 at 14:38
@DonAntonio: It is assumed that $P$ is normal in $G$. Also $p$ is not 2. – N K Jul 22 '12 at 15:20
Yikes! So there are LOTS of restrictions, and besides all these the order of G must be even...! Do you have some examples of such groups, to begin with? – DonAntonio Jul 22 '12 at 17:09
Could you at least provide some motivation for this question? – Derek Holt Jul 22 '12 at 17:16

I believe the semidirect product of an elementary abelian group of order 25 with a dihedral group of order 6 and faithful action is a counterexample. This is SmallGroup(150,5) in the GAP or Magma library.

There are 6 subgroups of order 5, 15 of order 2 and 15 cyclic subgroups of order 10. Note that 15 is not a multiple of 6.

More generally, if $q = (p+1)/2$ is odd, then I would expect the semidirect product of elementary $p^2$ with dihedral $2q$ (and faithful action) to be a counterexample. You would have a better chance if you asked whether $t_{2p}$ was a multiple of $(p+1)/2$.

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