Can anyone prove that there are only 2 non-isomorphic groups of order 6.
Every group has generators:
You can start by showing that G is either
1.generated by an element of order 6
2.an element of order 3 and an element of order 2.
Since all orders of the generators divide the group-order
(This follows from the Lagrange Theorem: #[G/H].#[H] = #G
For any subgroup H)
Once you figured that out you show that (1) and (2) are not isomorphic.
Secondly you show that 2 groups G and G' of type (1) are isomorhpic
and 2 groups of type (2) are isomorphic.
(This isomorphism is made by sending generators to generators)
suppose G is a group of order 6 and let $\displaystyle x,y \in G$ be with $\displaystyle o(x)=3, \ o(y)=2.$ since $\displaystyle [G:<x>]=2,$ we have $\displaystyle <x> \lhd G.$ now consider two cases:
1) $\displaystyle xy=yx$: in this case $\displaystyle o(xy)=6=|G|$ and thus $\displaystyle G = <xy> \cong C_6.$ where $\displaystyle C_6$ is the cyclic group of order 6.
2) $\displaystyle xy \neq yx$: since $\displaystyle <x> \lhd G,$ we have $\displaystyle yxy^{-1}=x^k,$ for some $\displaystyle k=0,1,2.$ if $\displaystyle k=0,$ then $\displaystyle x=1_G,$ which is not true. if $\displaystyle k=1,$ then $\displaystyle xy=yx,$ which is
the first case. so $\displaystyle k=2$ and $\displaystyle yxy^{-1}=x^2=x^{-1}.$ therefore $\displaystyle G \cong D_6,$ the dihedral group of order 6.