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In finite $p$-groups, the number of subgroups of order $p^k$ is congruent to $1 \mod p$.

Is it true that the number of normal subgroups of order $p^k$ is congruent to $1 \mod p$?

  • 8
    Well a conjugacy class of non-normal subgroups in a finite $p$-group has order divisible by $p$.2012-02-28
  • 0
    I am not sure if your first statement is right without $k$ being the maximal power of $p$ in $|G|$.2012-02-28
  • 1
    In any finite group of order divisible by $p^k$ with $p$ prime, the number of subgroups of order $p^k$ is congruent to 1 mod $p$.2012-02-28

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