Relation between isomorphism and commutative property Consider a mapping $f : G_1 \to G_2$ between two groups.   
I was told that if one group is abelian and the other is not abelian, we won`t possible be able to find a bijective homomorphism between them, concluding that they are not isomorphic.   
I was trying to understand formally ( by proof ) or intuitively why it is true .
 A: Let $G_1$ abelian, suppose you can find an iso $f:G_1\rightarrow G_2$. Take $b_1,b_2\in G_2$ arbitrary elements. By surjectivity of $f$, there exist $a_1,a_2\in G_1$ such that $f(a_i)=b_i$. Now you get:
$$b_1b_2=f(a_1)f(a_2)=f(a_1a_2)=f(a_2a_1)=f(a_2)f(a_1)=b_2b_1$$
which is abelianity of $G_2$. Hence if $G_2$ is supposed to be non-abelian, you can't find such an iso.
Actually you can't even find a surjective morphism
A: Assume $G_1$ is not abelian. Then there are $a,b\in G_1$ with $ab\ne ba$. As $f$ is injective, this implies that $f(a)f(b)=f(ab)\ne f(ba)=f(b)f(a)$, i.e. $G_2$ is not abelian.
Assume $G_2$ is not abelian. Then there are $a,b\in G_2$ with $ab\ne ba$. As $f$ is surjective, there are $c,d\in G_1$ with $f(c)=a, f(d)=b$.
Then $cd=dc$ would imply $ab=f(c)f(d)=f(cd)=f(dc)=f(d)f(c)=ba$.
A: Assume $G_1$ is abelian.  Also lets say that $x, y \in G_1$ are two elements and $f\colon G_1 \to G_2$ is a homomorphism.  Then
$$f(x)f(y) = f(xy) = f(yx) = f(y)f(x)$$
so the elements $f(x), f(y) \in G_2$ commute.  If $f$ is bijective (actually we only need surjective) then this means all the elements of $G_2$ commute, so $G_2$ is abelian.
So any group isomorphic to an abelian group is also abelian.  This should jive with our intuition of isomorphic groups being "essentially the same".  This also means if one group is abelian and the other isn't then they can't be isomorphic.  For them to be isomorphic they have to both be abelian or both non-abelian.
