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Hello all I have a question on group theory.

Let $G$ be any group and $H$ a finite normal subgroup of $G$. Suppose that the quotient $G/H$ is abelian. Is it true then that $G$ is abelian? If not, do you have a counterexample?

My attempt: I think the answer is positive, as taking only a finite piece of $G$ does not affect very much on its behavior.

Thanks in advance!

*Edit: I edit my question as I received counterexamples with finite groups (which I really appreciate). What if we add the assumption that $G$ is infinite?

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  • $\begingroup$ $G=S_3$, $H$ cyclic of order $3$ already gives you a counterexample. (note: I assume you meant to add the assumption that $H$ is abelian as well? Otherwise just take $G=H$ to generate cheap counterexamples.) $\endgroup$ – lulu May 1 '18 at 12:30
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    $\begingroup$ The direct product of an infinite abelian group with a finite nonabelian group $H$ is still a counterexample. So perhaps you want to assume that $H$ is abelian? (The answer is still no, but it is not quite so easy to think of examples.) $\endgroup$ – Derek Holt May 1 '18 at 12:43
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Counter-example: Let $S_n$ be the permutations of $n$ elements and $A_n$ be the even permutations. $A_n$ is a normal subgroup of $S_n$ and their quotient is of order 2, so Abelien, but $S_n$ is about as un-Abelian as it gets.

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As a counterexample, let $G$ be the smallest nonabelian group, the symmetric group on $3$ elements, $S_3$. Let $H = A_3$, the alternating subgroup of $S_3$. Notice $|H| = 3$, and $|G:H| =2$, so $H$ is a normal subgroup of $G$. The quotient is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, so is abelian.

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No. You can always find such a subgroup $H$ (assuming you refer to finite groups at least). So if that were true, it would follow that every group is abelian. The $H$ I have in mind is $H = G'$.

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Let me quickly expand on Erik Holts comment, that there are examples where both $H$ and $G/H$ are Abelian but $G$ is not: they are the dihedral groups $D_n$. Geometrically they are the symmetry groups ($n$ rotations and $n$ reflections) of the regular $n$-gon. Algebraically they are generated by elements $\rho$ and $\sigma$ where $\rho^n = 1, \sigma^2 = 1$ and $\rho^k \sigma = \sigma \rho^{n-k}$ for every $k$. The normal subgroup $H$ consists of the powers of $\rho$, geometrically this is the subgroup of only rotations, it is a cyclic group and hence abelian. The quotient has only two elements and hence is abelian as well.

The algebraic description makes clear how to make an infinite example: for every $k \in \mathbb{Z}$ we have an element $\rho^k$ and an element $\sigma \rho^k$ and multiplication is covered by $\rho^k \sigma = \sigma \rho^{-k}$ and of course $\sigma^2 = 1$.

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    $\begingroup$ These aren't the only examples. Consider $\mathbb{Z}_7 \rtimes_\varphi \mathbb{Z}_3$, with $\varphi: \mathbb{Z}_3 \rightarrow \mathrm{Aut}(\mathbb{Z}_7)$ an action by conjugation giving the presentation $\langle a,b \mid a^3, b^7, aba^{-1}b^{-2} \rangle$. This handles the set of non-direct product groups of order $pq$ for primes $p \neq q$ with $p \mid q-1$. Read more? There are other nonabelian ways to combine abelian groups, see the extension problem. $\endgroup$ – Eric Towers May 1 '18 at 20:45

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