I have a relatively complicated group, I will not go into detail about what it is, it is a group of automorphisms, and the group-relation is composition, so it is kind of complicated.
However, I am supposed to prove that this is ismorphic to $S_3$. In $S_3$, we have six elements, $a=id,b=(1~~2~~3),c=(1~~3~~2),d=(1~~2),e=(1~~3),f=(2~~3)$.
We have that $b$ is of order $3$, $d$ is of order $2$, and $b$, $d$ generate the group like this: $b^2=c,\ bd=e,\ db=f$.
Let's say that, in my complicated group, I identify one element $b'$ with order $3$, one element $d'$ with order $2$ (and elements, $c',e',f'$) with $b'^2=c',\ b'c'=e',\ d'b'=f'$.
Have I then showed that my group is isomorphic to $S_3$?
The problem with constructing a function and checking that it is a homomorphism (in order to be an isomorphism) is that I will have to check $72$ cases. So I tried this instead. But I am not sure if it is enough, could there be a group of 6 elements, with these properties, which is not isomorphic to $S_3$?
Claim: Any group of $G$ order 6 is $\mathbb{Z}/6$ or $S_3$. Thus, to check that a group of order 6 is $S_3$, it suffices to show that it is non-abelian.
Proof of claim: by Cauchy's theorem, $G$ has an element $g$ of order $2$ and an element $h$ of order $3$. The subgroup generated by $h$ has index two, so is normal. Thus $G$ is the semi-direct product of the cyclic subgroups $<g>$ and $<h>$, since these two groups must generate all of $G$ by a cardinality argument. There are precisely two such semi-direct products, corresponding to the trivial and nontrivial maps $$ \mathbb{Z}/2 \to (\mathbb{Z}/3)^\times. $$