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Prove that a group of order $2^{67}$ has a normal subgroup of order $2^{59}$.

I think we need to use the fact that it's a p-group, and then use the quotient group with the center here, but I am not sure how to approach the problem from here. Also, we can't use Sylow theorem, because it is beyond what we've learned.

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  • $\begingroup$ look up nilpotent groups. $\endgroup$ – Lubin Feb 24 '18 at 5:33
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All you need to know is a $p$-group has nontrivial center, and the fact that an abelian $p$-group has subgroups of all possible order (see below).

A group of order $p^n$ has a normal subgroup of order $p^k$ for any $0\leq k \leq n$.

Use induction on $n$, the base case $n=1$ is obviously true. Let $G$ be a group of order $p^n$, $Z$ be its center. If $G$ is abelian, we're done.

Now assume $G$ is not abelian, denote $|Z| = p^r$. Let $\pi: G\to G/Z$ be the canonical surjection. Since $r<n$, by induction hypothesis, for any $k\leq r$, $Z$ has a normal subgroup of order $p^k$. It is easy to verify this is also a normal subgroup in $G$. For $k>r$, by induction hypothesis on $G/Z$, we have $H \lhd G/Z$ with $|H| = p^{k-r}$, then $\pi^{-1}(H)$ is a normal subgroup of $G$ with order $p^k$.


To prove that an abelian $p$-group has subgroups of all possible order, one can proceed as follows: Let $G$ an abelian $p$-group, $|G| = p^n$, if $G$ is cyclic, then done. If $G$ is not cyclic, select $a\neq 1$, let $K$ be the group generated by $a$. Then $|K| = p^r$, $1\leq r \leq n-1$, use the same kind of induction as above applied to $K$ and $G/K$ respectively.

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  • $\begingroup$ How to prove than an abelian $p$-group has subgroups of all possible order? $\endgroup$ – jimmaal Feb 24 '18 at 6:04
  • $\begingroup$ @jimmaal I edited my answer. $\endgroup$ – pisco Feb 24 '18 at 6:26

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