Is this enough to prove that the group is isomorphic to $S_3$? I have a relatively complicated group, I will not go into detail about what it is, it 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 $S_3$?
 A: 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
$$
\mathbb{Z}/2 \to (\mathbb{Z}/3)^\times.
$$
A: I'm guessing that you have some Galois group, but if it is a finite extension then the Galois group is clearly a subgroup of the symmetric group on the roots of the minimal polynomial that generates the extension, since its action on the roots completely determines its action on other elements of the extension. Also, you can prove that if a subgroup of $S_p$ has a $2$-cycle and a $p$-cycle where $p$ is a prime, then the subgroup is $S_p$ itself.
