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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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    @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

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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 $J$ is normal in $G$?2013-05-03