Image of group morphism is abelian implies domain is abelian

Question

If $f$ is a non-trivial group morphism and if for all $g_1,g_2$, we have $f(g_1)f(g_2) = f(g_2)f(g_1) $ i.e. image is abelian. Does that imply $g_1g_2 = g_2g_1$ i.e. the domain is abelian?

I have been unable to prove this statement, so I thought I would construct a counter-example. However, to do that would require that my domain be a non-abelian group (for example, some $S_n$ or dihedral group) - however, I don't have much proof-related experience in constructing group morphisms yet, so I was unable to do that. Any ideas?

Answer 1

Take any two groups A and B, with A non-abelian and B abelian and non-trivial, and consider the projections $A\times B\to B$.

Answer 2

Consider

$$\begin{align} \varphi: S_3&\to \Bbb Z_2,\\ r^is^j&\mapsto s^j, \end{align}$$

where $S_3=\langle r,s\mid r^3, s^2, srs=r^{-1}\rangle $.

Answer 3

Unless I misunderstand your question, there is no reason that $f(g_1)f(g_2) = f(g_2)f(g_1)$ implies commutativity in the domain group.

For example take $G = \{$ similarities in $\mathbb{R}^3 \}$, and $f: G \to \mathbb{R}^+$ where $f(s) =$ similarity ratio of $s$.
$f$ is a group morphism from $(G, \circ)$ to $(\mathbb{R}^+, \times)$. However $(\mathbb{R}^+, \times)$ is abelian but not $(G, \circ)$.

Answer 4

Let $\Bbb Q_8$ be the quaternions. All it's subgroups are normal. Take any non-trivial one. Say $H=\{\pm1\}$.

Then do the canonical projection: $\Bbb Q_8\to \Bbb Q_8/H$.

$\Bbb Q_8$ is not abelian. But since the quotient has order $4$, it is.