Every homomorphic image of abelian group is abelian but converse need not be true.

Question

Every homomorphic image of abelian group is abelian but converse need not be true.

My exercise is to give an counterexample that converse need not be true.

My attempt : I tried to construct a mapping $f : S_3 \rightarrow \frac{S_3}{A_3}$ defined by $f(x) = x$ where $ x \in S_3$

I thinks this will be counter example but here $S_3$ is non- abelian.

Is jt true?

Answer 1

I would have written $f(x)=xA_3$, as that's what the elements of $S_3/A_3$ look like. But otherwise, yeah, that's a good example. This $f$ (and all other homomorphisms constructed from a group to a quotient of that group in this way) is called the canonical quotient homomorphism. Because it's the most obvious way to map a group into a quotient.

For an even simpler example, take any non-abelian group and map it to the trivial group (or to the identity element of your favourite group).

Answer 2

A well known example is the determinant mapping from the invertible $n\times n$ matrices over $\mathbb{R}$ ($GL(n,\mathbb{R})$) to the multiplicative group of $\mathbb{R}\setminus \{0\}$

$$\det: GL(n,R) \rightarrow \mathbb{R}\setminus \{0\}$$

Answer 3

Another counter example from trancelocation answer . $$\det: GL(n,R) \rightarrow \{-1,1\}$$