Let $\phi : G → K$ be an isomorphism of groups. Prove that $G$ is abelian if and only if $K$ is abelian.

Question

Let $\phi : G → K$ be an isomorphism of groups. Prove that $G$ is abelian if and only if $K$ is abelian.

I know I need to start with proving. For any elements $a,b \in G$, $a$ and $b$ commute iff $\phi(a)$ and $\phi(b)$ commute.

Answer 1

Let $G$ be abelian and $a,b \in K$, now there exist some $x,y \in G$ ,s.t. $\phi(x) = a, \phi(y) = b$, then $ab = \phi(x)\phi(y) = \phi(xy) = \phi(yx) = \phi(y)\phi(x) = ba$

Now we have that if $K$ is abelian and $x,y \in G$. $xy = yx \iff \phi(xy) = \phi(yx) \iff \phi(x)\phi(y) = \phi(y)\phi(x)$, but as they are elements in $K$ this is true.

Answer 2

Note that $$\phi(a)\phi(b)=\phi(b)\phi(a)\underbrace{\iff}_{(1)} \phi(ab)=\phi(ba)\underbrace{\iff}_{(2)} ab=ba,$$ where we use in $(1)$ the definition of homomorphism and in $(2)$ that $\phi$ is an isomorphism.

Answer 3

f(a)f(b)=f(b)f(a) iff f^-1 (f(a) f(b)) = f^-1 (f(a)f(b)) (by 1-1) iff ab=ba. Since f is surjective this is sufficient.