# Thread: Help with Lagrange's theorem.

1. ## 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

2. 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
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 .

3. 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.

4. 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!