# Help with Lagrange's theorem.

• Jun 4th 2010, 10:56 PM
Mai
Help with Lagrange's theorem.
Lagrange's Theorem states that: 'If H is a sub-group of a finite number G, then the order of H is a factor of G'.
It follows that 'The period of an element is a factor of the order of the group.'
Use the two theorems above and you knowlegde of groups to justify the proposition that only 2 groups of the order 6 exists, one Abelian and one non-Ablien. (please DON'T use cosets, and do use identities and subgroups or periods)

Provision of the supporting arguments in the form of proof :)
Thanks (Rofl)
• Jun 4th 2010, 11:45 PM
simplependulum
Quote:

Originally Posted by Mai
Lagrange's Theorem states that: 'If H is a sub-group of a finite number G, then the order of H is a factor of G'.
It follows that 'The period of an element is a factor of the order of the group.'
Use the two theorems above and you knowlegde of groups to justify the proposition that only 2 groups of the order 6 exists, one Abelian and one non-Ablien. (please DON'T use cosets, and do use identities and subgroups or periods)

Provision of the supporting arguments in the form of proof :)
Thanks (Rofl)

This is not the answer but opinions :

I read an essay from Arthur Cayley , the only three groups of order six are :

$1, \alpha , \alpha^2 , \alpha^3 , \alpha^4 , \alpha^5~~ ( \alpha^6 =1)$ and

$1,\alpha , \beta , \beta^2 , \alpha \beta , \alpha \beta^2 ~~ (\alpha^2 = 1) ~,~ (\beta^3 = 1)$

But there are two possible groups :

$\beta \alpha = \alpha \beta$ (commute)

$\beta \alpha = \alpha \beta^2$

however , the first possible group is isomorphic to the group first mentioned with $\alpha^6 = 1$ because

$\{\ 1 , \alpha \beta , \beta^2 , \alpha , \beta , \alpha \beta^2 \}\$

$= \{\ 1 , \alpha \beta , ( \alpha \beta)^2 ,( \alpha \beta)^3,( \alpha \beta)^4,( \alpha \beta)^5 \}\$

( $\alpha$ and $\beta$ commute )

Therefore , we get only two groups , they are :

$1, \alpha , \alpha^2 , \alpha^3 , \alpha^4 , \alpha^5~~ ( \alpha^6 =1)$

$1,\alpha , \beta , \beta^2 , \alpha \beta , \alpha \beta^2 ~~( \alpha^2 = 1) ~,~ (\beta^3 = 1 )~~ \beta \alpha = \alpha \beta^2$

one is abelian while the other is not .
• Jun 5th 2010, 03:32 AM
Mai
Hello simplependulum Thanks for the answer/opinions.
I am in grade 11 and Can you please roughly just very roughly explain the answer or prove for this question. (Nod)
• Jun 5th 2010, 07:04 AM
roninpro
You can break this up into two cases:

Case one: There is an element $g$ of order / period 6. Then, $g$ generates the group. Therefore, the group is cyclic and is isomorphic to $\mathbb{Z}_6$.

Case two: There are no elements of order / period 6. This means that the only possible orders / periods are 1, 2, 3, by Lagrange's Theorem. Suppose $g$ is an element of order 3. It forms a subgroup consisting of $g, g^2, g^3=e$. Take note that $g^2$ has order 3 as well. This means that order / period 3 elements come in pairs. Since there is one identity element, we only have 5 elements to "play with". Combining these two facts means that there can be only 2 or 4 order / period 3 elements. We will have trouble if there are 4 of them. (Try to prove this on your own!) This leaves 3 elements, all of which have to be order 2, by exhaustion.

In summary, we are forced to have 1 identity, 3 order / period 2, and 2 order / period 3 elements. Such a group is isomorphic to $S_3$, the group of permutations on three letters.

Best of luck!