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I thought this one would be easier than it has turned out to be. So given a group $G$ there exists some group $N\triangleleft G$ such that both $N$ and $G/N$ are abelian. I considered some arbitrary subgroup $K\subseteq G$ and then my instinct was to look at the group $K\cap N$ which is abelian in $G$ and both abelian and normal in $K$, but then I hit a snag looking at $K\;/\;K\cap N$.

If $K\subseteq N$ or $N\subseteq K$, then the proof is easy, but what happens when their intersection is not just $K$ or $N$, but smaller? possibly just the identity even. Neither the correspondence between subgroups of $G$ containing $N$ and subgroups of $G/N$, nor the 1st or 3rd isomorphism theorems seem to be of much help (the 2nd iso theorem is given as a later exercise and thus shouldn't be needed).

Can anyone help me with this? Thanks.

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Tiny comment: Why do you say 'abelian in $G$'? – Tara B Oct 20 '12 at 0:41
I dunno just to be thorough I guess, it is isn't it? – esproff Oct 20 '12 at 1:14
Well, but being abelian is a property of the (sub)group, which doesn't in any way depend on any overgroup (unlike normality), so I don't know what being abelian 'in $G$' should even mean. – Tara B Oct 20 '12 at 10:43
up vote 3 down vote accepted

Your instinct was good.

$K/(K\cap N)\cong KN/N$ ($n$-th isomorphism theorem, where $n$ is $2$ or $3$, I don't remember), and is therefore isomorphic to a subgroup of the abelian group $G/N$, and is hence abelian.

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This is the second isomorphism theorem, so this should be provable with out it, I'm not sure if I can use it or not. – esproff Oct 20 '12 at 1:15
Why wouldn't you be able to use it? – Tara B Oct 20 '12 at 10:41
nm, it's fine thanks. – esproff Oct 21 '12 at 17:10

Hint. Show that $G$ is metabelian iff $G''=1$ and that $H\le G$ implies $H'\le G'$.

The factor group $G/N$ is abelian iff $1=(G/N)'=[G/N,G/N]=[G,G]N/N$ iff $G'\le N$, hence there exists an abelian such $N$ iff $G'$ is abelian (for subgroups inherit abelianness and $G'$ abelian implies $N=G'$ is viable), iff $G''=1$. Now $H\le G$ entails $H'\le G'$, so it also entails $H''\le G''$, and subsequently if $G''=1$ then $H''=1$ so $H$ is metabelian if $G$ is metabelian.

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