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 October 11th, 2017, 11:58 PM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 Number of p-subgroup I am reviewing for my exam in algebra and i need help to answer the following problem: >Let $G$ be a finite group, $p$ a prime number that devides $|G|$ and suppose that $G=p^kn$ with p doesn't devides $n$. For $1\le j\le k$, let $m_j(G)$ be the number of subgroups of $G$ of order $p^j$. >Suppose that $p^2$ devides $|G|$. Let $X=\left\{(H,K)|HShow that$|X|=m_1(G) \pmod p$and$|X|=m_2(G) \pmod p$and conclude that$m_2(G)=1 \pmod p\$ Thanks in advance for your help.

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