# Groups of Order 6

• Dec 16th 2009, 07:12 PM
Groups of Order 6
Can anyone prove that there are only 2 non-isomorphic groups of order 6.
• Dec 16th 2009, 07:49 PM
Drexel28
Quote:

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.
• Dec 16th 2009, 08:05 PM
NonCommAlg
it's a nicer problem to prove that for any prime number $\displaystyle p$ there are only two groups (up to isomorphism) of order $\displaystyle 2p.$ (Nod)
• Dec 17th 2009, 04:56 AM
...

that for any prime number http://www.mathhelpforum.com/math-he...97a8c47a-1.gif there are only two groups (up to isomorphism) of order http://www.mathhelpforum.com/math-he...7c8b6f69-1.gif
• Dec 18th 2009, 11:39 AM
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)
• Dec 18th 2009, 11:49 AM
Swlabr
Quote:

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.
• Dec 18th 2009, 12:17 PM
NonCommAlg
Quote:

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 $\displaystyle x,y \in G$ 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.