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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?

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  • $\begingroup$ Suppose you took the trivial homomorphism $\phi(a,b,c,d)=1$. That would not help! $\endgroup$
    – almagest
    Jan 30, 2018 at 17:01
  • $\begingroup$ @almagest Would $\phi(a,b,c,d) = 1$ satisfy $\phi((a,b,c,d)\circ(e,f,g,h)) = \phi(a,b,c,d) \phi(e,f,g,h)$? $\endgroup$
    – Meik Vtune
    Jan 30, 2018 at 17:04
  • $\begingroup$ Why not? $1\times 1=1$. $\endgroup$
    – almagest
    Jan 30, 2018 at 17:06
  • $\begingroup$ I won't speak about homomorphisms in general, but you should be more explicit about $M(2, 2)$. In particular... does $\pmatrix{2 & 0 \\ 0 & 1 }$ have an inverse in $M$? $\endgroup$
    – pjs36
    Jan 30, 2018 at 17:10
  • $\begingroup$ @pjs36 I see what you mean, using M twice wasn't smart, but if we ignore that, $M$ clearly only holds elements of $M(2,2)$ where $\det(A) = 1$, thus the inverse exists. $\endgroup$
    – Meik Vtune
    Jan 30, 2018 at 17:12

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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.

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  • $\begingroup$ Could you elaborate a little bit on why a homomorphism alone won't work? That's the real point I fail to understand. The only way I've been able to explain it to myself is that if $f$ is bijective, we can calculate whatever we want in the known group and use the inverse $f^{-1}$ to return back to our original set/group. $\endgroup$
    – Meik Vtune
    Jan 30, 2018 at 17:54
  • $\begingroup$ As I said, it was already answered in the comments. A homomorphism $f:N\rightarrow M$ might not give any non-trivial property about $M$. Take the trivial homomorphism $f$, mapping everything to the matrix $I_2$ in $M$. So we only have then the identity matrix $I_2$, i.,e., the image is the trivial group. $\endgroup$
    – Dietrich Burde
    Jan 30, 2018 at 19:11

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