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I have a group $G$ of order $p^n$ for $n \ge 1$ and $p$ a prime. I am looking for two specific subgroups within $G$: one of order $p$ and one of order $p^{n-1}$. I don't think I would use the Sylow theorems here because those seem to apply to groups with a "messier" order than simply $p^n$. Would Cauchy's Theorem allow me to generate the two requisite subgroups? I could use it to find an element of order $p$ and an element of order $p^{n-1}$ and then consider the cyclic subgroups generated by these two elements?

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Your idea is good for the subgroup of order $p$. But you can't use Cauchy's theorem to get an element of order $p^{n-1}$ (and no such element will exist in many cases). –  mt_ Feb 17 '13 at 17:27
    
Then letting $a \in G$ be an element of order $p$, perhaps I should consider $G/\langle a \rangle$? –  Zvpunry Feb 17 '13 at 17:32
    
@JJR $\langle a\rangle$ need not be normal for arbitrary $a$. –  Hagen von Eitzen Feb 17 '13 at 17:32
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2 Answers

up vote 11 down vote accepted

Hints:

1) Use the class formula and deduce that $\,|Z(G)|>1\,$

2) Use induction now to show that $\,G\,$ has a normal subgroup of order $\,p^k\,\,,\,\,\forall\,\,0\le k\le n\,$

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Looks tractable. Thanks for your help! –  Zvpunry Feb 17 '13 at 17:36
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Let $P$ act on itself by conjugation. $1$ appears in an orbit of size $1$, and everything else appears in an orbit of size $p^k$ for some $k$. Since the sum of the orbit sizes is equal to $|P|$, which is congruent to $0\mod{p}$, that means there has to be at least one more orbit of size $1$. Orbits of size $1$ under conjugation contain elements which commute with everything in the group; they compose $Z(P)$. Now suppose inductively that $\exists S\unlhd P$ with $|S|=p^k$. Then by the above lemma and Cauchy's theorem $P/S$ has a central subgroup $\overline{Q}$ of order $p$. Lifting $\overline{Q}$ back to $P$, we obtain a normal subgroup of order $p^{k+1}$.

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