If $G$ is a finite group, $|G| = p^n$ , $p$ is a prime, then how do I prove $G$ has a subgroup of order $p^k$, for each $k$, $1 \leq k \leq n$?
Subgroups of $G$, $|G| = p^n$
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12By induction. It is enough to show that if $n>1$ then $G$ has a proper non-trivial normal subgroup (and you probably know that $G$ has a non-trivial center, so you can use that) – 2011-09-24
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0Did you try the technique I suggested in a comment to your [other question](http://math.stackexchange.com/questions/67028/g-pn-p-is-a-prime-and-h-is-subgroup-of-g)? – 2011-09-24
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0Yeah.By induction hypothesis $G/Z(G)$ of order $p^m$ must subgroups for order $p^k$ for each $K, 1 \leq k \leq m$.So,$G$ should have subgroups of order $p^k$, for each $k, 1\leq k\leq n$. Is this correct? How do I prove the abelian case? – 2011-09-24
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1The abelian case you can deal with as in Patrick's answer, for in that case the subgroup $H$ he construct *is* normal. – 2011-09-24
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0@user774025: You need to explain how you get the subgroups of $G$ from subgroups of $G/Z(G)$. You've just asserted it but not proven it. – 2011-09-25
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0That is because there's one to one correspondence between subgroups of $G/Z(G)$ and subgroups of $G$ containing $Z(G)$. – 2011-09-25
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0@user774025: Yes, but subgroups of size 1 through $p^m$ in $G/Z(G)$ become subgroups of size $p^{n-m}$ through $p^n$ in $G$ under this correspondence. How do you get subgroups of size 1 through $p^{n-m}$ in $G$? (Sorry if you've figured out all this already and think I'm being a little pedantic, but I think this should be stated in a complete proof.) – 2011-09-25
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0$Z(G)$ is of order $p^{n-m}$. So, by induction hypothesis $Z(G)$ should have subgroups of order 1 through $p^{n-m}$ and this subgroups are also subgroups of $G$. Right? – 2011-09-25
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1@user774025: Yes, you got it! If you want another exercise of this form, try proving the stronger result that in a group $G$ of order $p^n$, there always exists a chain of subgroups $G_0 < G_1 < \ldots < G_n$ where $G_i$ has order $p^i$, *and* every $G_i$ is *normal* in $G$. – 2011-09-25
1 Answers
For induction, this result is trivial for $n=1$, and it is more easy to show that the subgroups you're looking for are normal.
Use Cauchy's theorem to note that there is an element of order $p$ in the center of $G$. Call $\langle p \rangle = H$. This subgroup is normal. Thus $G/H$ has order $p^{n-1}$, and by induction hypothesis there exists normal subgroups of $G/H$ of order $p^k$ for $1 \le k \le n-1$ in $G/H$, and those subgroups of $G/H$ are giving you a normal (in $G/H$) subgroup $J/H$ for some subgroup $J$ of $G$. Hence $J$ has order $p^k$ for $2 \le k \le n$ and is normal in $G$. For the case $k=1$ here, you can just take $H$ since $|H|=p$.
Note that I was a little foggy about the apparition of $J$... you need to be more careful than I was, but I'll let you fill in the details.
Hope that helps,
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0why is the subgroup H normal? – 2011-09-24
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0It isn't. Pick any non-abelian subgroup of order $p^3$: most elements do not generate normal subgroups. – 2011-09-24
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1Pick an element of order $p$ in the center, and it should work. – 2011-09-24
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0why $J$ is normal in $G$? – 2013-05-03