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Let $|G| = pq^n$ for $p primes and $n$ is in the natural numbers

a) Show there is $H$ a normal subgroup of $G$ with $|H|=q^n$

b) If $P$ is a normal subgroup of $G$ with $|P| = p$, show that for any $m$ with $m$ divides $|G|$, there is $H_m$ (a subgroup of $G$) with $|H_m| =m$.

I am completely lost on how to start this, so any help would be appreciated, thanks!

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    yes p2012-12-18

2 Answers 2

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Hints:

a) You know such a subgroup exists by (BLANK)'s theorems, and it's normal since its index is the smallest (BLANK) dividing the order of the group.

b) Since you know that $H$ (the group from a)) and $P$ are normal, their product $PH$ is $G$, and their intersection is trivial (why?) you know that $G$ is isomorphic to their (BLANK). Now, the property you want holds true for $p$-groups (this is a common fact) and it should hold true for (BLANKS) of $p$-groups as in this problem by just considering the problem (BLANK)wise.

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Hints:

1) Show $\,G\,$ is a direct product of prime-power order groups (for this you need both to prove (a) and the given info in (b))

2) Show that for any divisor of the order of a $\,p$-group (in general, with $\,p\,$ a general prime) there exists a subgroup of order that divisor. In fact, we can sharpen this by adding "normal" before the word subgroup. You may want to use induction and the fact that the center of any finite $\,p-$group is non-trivial