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

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- Dec 16th 2009, 07:12 PMpleasehelpme1Groups of Order 6
Can anyone prove that there are only 2 non-isomorphic groups of order 6.

- Dec 16th 2009, 07:49 PMDrexel28
- Dec 16th 2009, 08:05 PMNonCommAlg
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 AMpleasehelpme1...
can your prove that then.

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 AMDinkydoe
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 AMSwlabr
- Dec 18th 2009, 12:17 PMNonCommAlg
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.