I imagine the following result is trivial, but I wanted to check with the Group to make sure that my intuition is correct.

Consider a Group Homomorphism between two Groups $G_{1} = (A, +)$, $G_{2} = (B, \times)$, i.e.

$g:A \rightarrow B$

Such that $\forall a,b \in A$

$g(a + b) = g(a) \times g(b)$

Now let's say we endow $G_{1}$ with commutativity, i.e.

$a + b = b + a$

Then if we map that under $g$ we find

$g(a + b) = g(b + a)$

$g(a) \times g(b) = g(b) \times g(a)$

And so we see that if $G_{1}$ is commutative then that property is inherited in the mapping onto $G_{2}$, i.e. if $g$ exists and $G_{1}$ is commutative then the set it maps onto does not necessarily have to be commutative but it only maps onto those elements in $G_{2}$ that are commutative!

I thought this was a cool result. Is this correct?


The result of your statement is true and can be simplified as below:

Let $G_1,G_2$ be groups and $g:G_1\rightarrow G_2$ be a homomorphism. If $G_1$ is abelian, then $g(G_1)$, the image of $G_1$ under the homomorphism $g$ is also abelian.

Bear in mind that this does not imply that $G_2$ is abelian.
For counterexample consider $\phi:\mathbb{Z}\rightarrow S_3$ the trivial homomorphism.

  • $\begingroup$ Thanks Alan, and appreciate the summary as your provided. That is what I was trying to say but didn't know the correct formation. I do love this property even it it's trivial. $\endgroup$ – user150203 Jul 25 '18 at 5:07

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