There are 2 groups of order 6 (up to isomorphism) I need to show that there are only 2 groups of order 6 up to isomorphism.
I did prove it, but the proof is quite cumbersome. I wonder if there is a very concise proof.
My proof outline: Suppose $G$ is of order 6 and is not $\mathbb{Z}_6$. Then any element that is not identity must have order 2 or 3.
I went on to show that there must be a element of order 2 and another one of order 3, and the intersection of the cyclic subgroups generated by the two is the identity. Then I'm able to show it is isomorphic to $S_3$.
 A: Suppose you have elements $a$ and $b$ of order $2$ and $3$ respectively. The six elements $a^rb^s: 0\le r \le 1, 0\le s\le 2$ are all distinct. If you then know which of these six elements is $ba$ then all the products are determined (by use of associativity).
If $ba=ab$ then the group is abelian and hence cyclic.
If $ba=b^s$ then $a=b^{s-1}$ and this does not work ($a$ and $b$ would then commute as in the first case). Similarly if $ba=a$ then $b=1$ contrary to hypothesis.
The only other possibility is $ba=ab^2$ and this gives a second group of order $6$ which can be realised as the group of symmetries of an equilateral triangle (dihedral group) or the symmetric group on three objects - to prove that such a group exists.

Another way of doing this is by counting elements. If there is a single subgroup of order $2$ then it is normal and any element of order $3$ commutes with the single element of order $2$. This gives the cyclic group.
Since there are five non-identity elements and elements of order $3$ come in pairs, there are two elements of order $3$ and three elements of order $2$ each of which generates a subgroup of order $2$.
You can show that the group acts transitively and faithfully by conjugation on these three subgroups of order $2$, which gives an isomorphism with $S_3$.
A: Let $G$ be a group of order $6$. By Cauchy's theorem, there exists some $H\subset G$ isomorphic to $\mathbb{Z}_3$. Conjugation gives a map $\rho:G \to S_2$ (identifying $S_2$ with permutations of the set of left cosets of $H$) with $\rho(H) = 1$. The only subgroups of $G$ containing $H$ are $G$ and $H$ themselves (as the index of such a group is divisible by $[G:H] = 2$). In the first case, $H$ is normal by definition; in the second case, it's the kernel of a map and thus normal.
Hence $H$ is normal in $G$. Choose some $g\in G$ with $g\not\in H$, and consider the action $\rho(g)\in \operatorname{Aut}(G/H) = \mathbb{Z}_2$ on $H$ by conjugation. If $\rho(g) = 1$, then $G$ is abelian and thus must be $\mathbb{Z}_6$. If $\rho(g) \not = 1$, then $G$ has presentation
\begin{align*}
G &= \left\langle g, h\;\Big\vert\,\; g^3 = 1;\, h^{-1}gh = g^2\right\rangle,
\end{align*}
which you can show directly is isomorphic to $S_3$ (or just note, with a bit more care, that we've shown that there's only one nonabelian group of order $6$, and $S_3$ is clearly such a group).
