1. Groups of Order 6

Can anyone prove that there are only 2 non-isomorphic groups of order 6.

Can anyone prove that there are only 2 non-isomorphic groups of order 6.
What knowledge do you have? What theorems do you know that can be used? I mean, if you are a masochist you can draw up the Cayley tables.

3. it's a nicer problem to prove that for any prime number $p$ there are only two groups (up to isomorphism) of order $2p.$

4. ...

that for any prime number there are only two groups (up to isomorphism) of order

5. 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)

6. Originally Posted by Dinkydoe
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)
In (2) you also want the generators to not commute with one another, as otherwise you get your cyclic group again.

7. Originally Posted by Swlabr
In (2) you also want the generators to not commute with one another, as otherwise you get your cyclic group again.
suppose G is a group of order 6 and let $x,y \in G$ be with $o(x)=3, \ o(y)=2.$ since $[G:]=2,$ we have $ \lhd G.$ now consider two cases:

1) $xy=yx$: in this case $o(xy)=6=|G|$ and thus $G = \cong C_6.$ where $C_6$ is the cyclic group of order 6.

2) $xy \neq yx$: since $ \lhd G,$ we have $yxy^{-1}=x^k,$ for some $k=0,1,2.$ if $k=0,$ then $x=1_G,$ which is not true. if $k=1,$ then $xy=yx,$ which is

the first case. so $k=2$ and $yxy^{-1}=x^2=x^{-1}.$ therefore $G \cong D_6,$ the dihedral group of order 6.