Abelianity and quotients

Question

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?

Answer 1

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.

Answer 2

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 $H = G'$.

Answer 3

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.

Answer 4

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 $\sigma^2 = 1$.