If I have the following:
$N := \{(a,b,c,d) \in \mathbb{Z}^4 \mid ad-bc = 1\}$
$(a,b,c,d) \circ (e,f,g,h) := (ae+bg,af+bh,ce+dg,cf+dh)$
The task is: Check $(N, \circ)$ for Group properties
I've found an Isomorphism to the following:
$M := \{A\in M(2,2) \mid \det(A) = 1\}$
with the standard matrix-product.
Let $f:N \to M$ with $(a,b,c,d) \mapsto \begin{bmatrix}a & b\\c & d\end{bmatrix}$
$f$ is bijective, which is fairly obvious.
I've shown that $f$ does indeed keep the properties needed for an isomorphism, since that is not the point, I'll leave that out.
Now we know that
$(M,*)$ is a non-abelian group, so $(N,\circ)$ is also a non-abelian group.
Now that's straight forward, the question that came up was, can we prove the same without a bijective mapping and if not, why?
To make clear what I mean: Would a homomorphism between $N$ and $M$ prove the same thing?
A homomorphism alone would not be enough, as was already discussed in the comments. But perhaps even more important is to know natural examples of groups; and here the group of integral matrices with determinant $1$, $M=SL_2(\mathbb{Z})$ is much more natural than $N$, although they are of course isomorphic. So it really makes sense to use the bijection, and to understand its value. Closely related to $M$ is the modular group, which arises in several contexts of mathematics and physics.