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
1
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group-theory
finite-groups
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0Where 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... – 2012-07-22
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0$A_4$ is a counterexample if you allow $p = 2$, although presumably you want to disallow this. – 2012-07-22
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0@DonAntonio: It is assumed that $P$ is normal in $G$. Also $p$ is not 2. – 2012-07-22
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0Yikes! 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? – 2012-07-22
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0Could you at least provide some motivation for this question? – 2012-07-22
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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