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Let $G$ be a finite group such that Sylow $p$-subgroups of $G$ are cyclic of order $p^2$. I know the number of cyclic subgroups of order $p^2$ is equal to the number of Sylow $p$-subgroups. I would like to know what is the number of cyclic subgroups of order $p$?

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Every subgroup of order $p$ is contained in a Sylow $p$-subgroup. Every cyclic group of order $p^2$ contains exactly one subgroup of order $p$. The question is whether two Sylow $p$-subgroups have trivial intersection, or nontrivial. –  Daniel Fischer Sep 11 '13 at 14:30
    
How much of the theory of these groups do you know? Are you aware that the $p$-Sylow, where $p$ is the largest prime divisor, is normal? –  Tobias Kildetoft Sep 12 '13 at 7:00
    
@Tobias Kildetoft: If a Sylow $p$-subgroup is normal, then the number of Sylow $p$-subgroups is $1$. Then the number of cyclic subgroups of order $p$ in the $G$ is $1$. –  User1257 Sep 12 '13 at 8:17
    
So we now know that it is at least the case when $p$ is the largest prime divisor. Now consider the case where $|G|$ has only two prime divisors. Can it happen that the $q$-Sylows (where $q$ is the other prime divisor) intersects non-trivially? –  Tobias Kildetoft Sep 12 '13 at 8:22
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I think you can show that if the Sylow $p$-subgroups are cyclic of order $p^n$, then for $0 \leq k \leq n$, the number of subgroups of order $p^k$ is $\equiv 1 \mod{p^{n-k+1}}$. So in your question, the number of subgroups of order $p$ is $\equiv 1 \mod{p^2}$. –  Mikko Korhonen Sep 12 '13 at 18:05

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