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Let $G$ be a group of order $p^\alpha$, where $p$ is prime. If $H\lhd G$, then can we find a normal subgroup of $G/H$ that has order $p$?

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Not if $H=G$. Otherwise, of course: order of $G/H$ divides order of $G$, so it's a power of $p$, so it has an element of order $p$, done. –  Gerry Myerson Jun 24 '12 at 7:59
    
Ok, I mistyped my question. I was wondering if we can find a normal subgroup of $G/H$ that has order $p$. –  youngtableaux Jun 24 '12 at 8:10
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I saw somewhere, if $|G|=m$ and $p$ is the smallest prime number dividing the order of $G$ then every subgroup of $G$ with index $p$ is normal in $G$. As Gerry noted; $|\frac{G}{H}|=p^s$ for any $s$. I think by taking $m=p^s$ we can solve the problem. I hope it help. –  B. S. Jun 24 '12 at 8:24
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I repeat: not if $H=G$. Other than that, I don't think $G/H$ differs form any other $p$-group. Does every finite $p$ group have a normal subgroup of order $p$? If so, then the answer t your question is yes. If no, let $A$ be a $p$-group with no normal subgroup of order $p$, let $G=A\times B$ for some $p$-group $B$, then $1\times B$ is normal in $G$, etc. –  Gerry Myerson Jun 24 '12 at 8:56
    
It may help you to note that finite non-trivial $p$-groups have non-trivial centers. –  Geoff Robinson Jun 24 '12 at 9:47
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1 Answer

Theorem: a group $\,G\,$ of order $\,p^n\,,\,p\,$ a prime, $\,n\in\mathbb N\,$ , always has normal subgroup of order $\,p^m\,\,,\,\,\forall\, m\leq n\,\,,\,m\in \mathbb N$

Proof: Exercise, using that always $|Z(G)|>1\,$ and induction on $\,n\,$

So the comments by Gerry and Geoff close the matter.

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