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October 11th, 2017, 11:58 PM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1  Number of psubgroup
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 $X=\left\{(H,K)H<K<G, H=p,K=p^2\right\}$. >Show 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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