Question

If G is an abelian group and phi: G rightarrow G' is a homomorphism of G onto G', prove that G' is abelian.

Answer 1

If
`\phi` is a homomorphism, then we have, for every
`g,h \in G`

`\phi(gh) = \phi(g)\phi(h)`

Since G is abelian, we have
`gh=hg`, and thus
`\phi(gh)=\phi(hg)`

But we also have

`\phi(gh)=\phi(g)\phi(h)=\phi(h)\phi(g)=\phi(hg)`

which proves that G' is abelian.

In other words, for every
`x,y \in G'`, you have
`xy=yx`, because there exist
`g,h \in G` such that
`x=\phi(g),\ y=\phi(h)`, and you can think of
`xy` as
`\phi(g)\phi(h)`, and of
`yx` and
`\phi(h)\phi(g)`

Then, you observe that
`\phi(g)\phi(h)=\phi(h)\phi(g)` because they mean
`\phi(gh)=\phi(hg)`, and
`gh=hg` by hypothesis, because G is abelian.