I'm hoping that the title of this is not vague, but if it helps at all, I'll give a specific example of what I have in mind. This isn't a problem I need help on, per se, but I'm concerned about whether I fully understand the proof technique, which seems like a non-constructive proof of existence. Here is a paraphrased version of a proof that the groups $D_3$ (symmetries of an equilateral triangle) and $S_3$ are isomorphic.
$D_3$ permutes the vertices of a triangle, so identifying a symmetry with a permutation of the indices $\{1,2,3\}$, $D_3$ can be thought of as a subgroup of $S_3$. This induces an injective homomorphism from $D_3$ to $S_3$. As $|D_3| = |S_3| = 6$, this injection is also surjective and therefore bijective.
It is possible that my wording is a bit sloppy (if so, I would greatly appreciate feedback on that), but I am not certain that I understand exactly what the homomorphism is. I can certainly draw out the elements of $D_3$, determine the permutation of vertices, and then write down an element-by-element map. It even makes sense that with this identification, I can identify $D_3$ as a subset of $S_3$. It's a group itself, so that implies that it's closed, contains the identity, etc. My question is: how can I assert that this relationship "induces an injective homomorphism"? That is, how do I know that the composition laws agree in the sense that $f(xy) = f(x) f(y)$? The only way I could think to check this is by brute force, for which there are quite a bit of elements. It seems that every proof of this fact I read do not do this brute-force verification and refer to a shortcut as described above. I've seen similar arguments for showing that the group $S_4$ is isomorphic to the rotations of the cube.
With the geometric definition of the dihedral group it's kinda intrinsic that the permutation of the vertices induced by a symmetry $g$, followed by the permutation of the vertices induced by another symmetry $h$, is equal to the permutation of the vertices induced by the symmetry "$h$ after $g$", namely $hg$: $$h\cdot(g\cdot V)=(hg)\cdot V \tag1$$ With the property $(1)$ agreed, the map $\varphi\colon D_3\to S_V$ defined by $\varphi_g(V):=g\cdot V$ is indeed a group homomorphism. In fact, for every vertex $V$, $\varphi_{hg}(V)=$ $(hg)\cdot V\stackrel{(1)}{=}h\cdot(g\cdot V)=$ $\varphi_h(\varphi_g(V))=$ $(\varphi_h\varphi_g)(V)$, whence $\varphi_{hg}=$ $\varphi_h\varphi_g$.