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 $p+1$?
The number of cyclic subgroup
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group-theory
finite-groups
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0@Derek Holt: Let $t_{k}$ be the number of elements of order $k$. I know if $|P|=p$, then $t_{2p}$ is a multiple of $t_{p}$. In this case $|P|=p^{2}$ and the number of Sylow $p-$subgroups is $1$. If we can prove that the number of cyclic subgroup of order $2p$ is a multiple of $p+1$, then it conclude that $t_{2p}$ is a multiple of $t_{p}$. – 2012-07-23
1 Answers
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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 $(p+1)/2$.