Subgroups of $G$, $|G| = p^n$

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 $k$, $1 \leq k \leq n$?

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By 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) –  Mariano Suárez-Alvarez Sep 24 '11 at 3:59
Did you try the technique I suggested in a comment to your other question? –  Ted Sep 24 '11 at 5:05
Yeah.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? –  Mohan Sep 24 '11 at 5:20
The abelian case you can deal with as in Patrick's answer, for in that case the subgroup $H$ he construct is normal. –  Mariano Suárez-Alvarez Sep 24 '11 at 5:42
@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 $G$. –  Ted Sep 25 '11 at 19:08

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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why is the subgroup H normal? –  Mohan Sep 24 '11 at 4:51
It isn't. Pick any non-abelian subgroup of order $p^3$: most elements do not generate normal subgroups. –  Mariano Suárez-Alvarez Sep 24 '11 at 4:55
Pick an element of order $p$ in the center, and it should work. –  Soarer Sep 24 '11 at 5:29
why $J$ is normal in $G$? –  user42912 May 3 '13 at 2:02