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October 11th, 2017, 11:58 PM   #1
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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 $X=\left\{(H,K)|H<K<G, |H|=p,|K|=p^2\right\}$.

>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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