# Lagranges theorem

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• Jan 5th 2010, 03:00 AM
Lagranges theorem
I have tried to understand this using various websites and m,y notes but i still dont understand whats going on.

Could someone please explain what im meant to do to prove a a set in lagranges theorem thanks.
• Jan 5th 2010, 03:19 AM
Swlabr
Quote:

I have tried to understand this using various websites and m,y notes but i still dont understand whats going on.

Could someone please explain what im meant to do to prove a a set in lagranges theorem thanks.

What is it your are wanting us to explain how to prove?

Lagranges Theorem states that the order of every subgroup $\displaystyle H$ of a finite group $\displaystyle G$ divides the order of $\displaystyle G$, $\displaystyle H \leq G \Rightarrow |H| \mid |G|$.

This is, perhaps, a result which is easier to understand once you have played around with it a bit. So, pick a few of your favourite groups and verify the result.

For instance, the Klein 4-group has order 4 and subgroups of order 1, 2, 2, 2 and 4, and cyclic groups of prime order (for instance, $\displaystyle \mathbb{Z}/7\mathbb{Z}$) have no proper subgroups.
• Jan 5th 2010, 03:36 AM
Quote:

Originally Posted by Swlabr
What is it your are wanting us to explain how to prove?

Lagranges Theorem states that the order of every subgroup $\displaystyle H$ of a finite group $\displaystyle G$ divides the order of $\displaystyle G$, $\displaystyle H \leq G \Rightarrow |H| \mid |G|$.

This is, perhaps, a result which is easier to understand once you have played around with it a bit. So, pick a few of your favourite groups and verify the result.

For instance, the Klein 4-group has order 4 and subgroups of order 1, 2, 2, 2 and 4, and cyclic groups of prime order (for instance, $\displaystyle \mathbb{Z}/7\mathbb{Z}$) have no proper subgroups.

What i dont understand is how we calculate g and h
• Jan 5th 2010, 03:47 AM
Swlabr
Quote:

What i dont understand is how we calculate g and h

Do you mean how do you calculate the orders of the group and the subgroup? This is just the number of elements in your group.

For instance, the Klein 4-group has order 4 as it has precisely 4 elements, and the cyclic group of order 7 has, well, order 7. It is just the elements $\displaystyle \{0, 1, 2, 3, 4, 5, 6\}$ under addition modulo 7.
• Jan 5th 2010, 03:52 AM
HallsofIvy
Quote:

What i dont understand is how we calculate g and h

We don't!

G is a given group and H is a given subgroup of G. If by "g" and "h" you mean the number of elements in G and H, respectively, when we are given G and H, we are given g and h.

I think you should look closely at the concept of "left cosets" of H. They are crucial in Lagrange's theorem and important for other things as well. For any x in G, its left coset is the set {xy| y in H}. (A "right" coset would be of the form {gx| y in H}.) You should look at the proofs that
1) Every left coset contains the same number of elements.
(And since {1y |y in H} is just H itself, that is just h.)

2) Every member of G is in exactly one such left coset.

If there are, say, n left cosets, each containing h members, and each member of G is in exactly one, we must have g= nh so h divides g.
• Jan 5th 2010, 03:53 AM
Quote:

Originally Posted by Swlabr
Do you mean how do you calculate the orders of the group and the subgroup? This is just the number of elements in your group.

For instance, the Klein 4-group has order 4 as it has precisely 4 elements, and the cyclic group of order 7 has, well, order 7. It is just the elements $\displaystyle \{0, 1, 2, 3, 4, 5, 6\}$ under addition modulo 7.

In my notes it says If G is a finite group of order g = ¦G¦ and H is a subgroup of order h = ¦H¦, then h must be a factor of G.

So if the set was G = {1, -1, i, -i}, i = ((-1)^0.5). I know how to show its a group by the axioms.

But then the question says obtain a non-trivial solution subgoup H which i ahve no idea to do. And use it to illustrate lagranges theorem for a finite group.

Would G and H be 4 in this case?
• Jan 5th 2010, 03:57 AM
Swlabr
Quote:

In my notes it says If G is a finite group of order g = ¦G¦ and H is a subgroup of order h = ¦H¦, then h must be a factor of G.

So if the set was G = {1, -1, i, -i}, i = ((-1)^0.5). I know how to show its a group by the axioms.

But then the question says obtain a non-trivial solution subgoup H which i ahve no idea to do. And use it to illustrate lagranges theorem for a finite group.

Would G and H be 4 in this case?

A non-trivial subgroup is a subgroup which is not equal to the group itself nor the trivial group. Thus, it must have order strictly greater to one and not equal to that of the group. The group you have been given is the Klein 4-group which I havev mentioned above. It has 3 non-trivial subgroups each of order 2. Can you find them? (Hint: Pick an element. Any element...other than 1...)
• Jan 5th 2010, 04:01 AM
Quote:

Originally Posted by Swlabr
A non-trivial subgroup is a subgroup which is not equal to the group itself nor the trivial group. Thus, it must have order strictly greater to one and not equal to that of the group. The group you have been given is the Klein 4-group which I havev mentioned above. It has 3 non-trivial subgroups each of order 2. Can you find them? (Hint: Pick an element. Any element...other than 1...)

why cant it be one?

thanks for your help by the way it is much appreciated.
• Jan 5th 2010, 04:45 AM
Defunkt
Quote:

why cant it be one?

thanks for your help by the way it is much appreciated.

By definition, if $\displaystyle H$ is a non-trivial subgroup of $\displaystyle G$, then $\displaystyle H \neq G ~ \text{and} ~ H \neq \{1_G\}$ where $\displaystyle 1_{G}$ is the identity element of G and the group $\displaystyle \{1_G\}$ is called the trivial subgroup of G.
• Jan 5th 2010, 04:49 AM
Quote:

Originally Posted by Defunkt
By definition, if $\displaystyle H$ is a non-trivial subgroup of $\displaystyle G$, then $\displaystyle H \neq G ~ \text{and} ~ H \neq \{1_G\}$ where $\displaystyle 1_{G}$ is the identity element of G and the group $\displaystyle \{1_G\}$ is called the trivial subgroup of G.

well -1, i and -i cant be in it as well then?
• Jan 5th 2010, 04:56 AM
Swlabr
Quote:

well -1, i and -i cant be in it as well then?

Why not?
• Jan 5th 2010, 04:57 AM
Defunkt
• Jan 5th 2010, 04:58 AM
Quote:

Originally Posted by Swlabr
Why not?

because they are in G
• Jan 5th 2010, 05:05 AM
Swlabr
Quote:

because they are in G

A subgroup is a set of elements from G which form a group under the operation of G.

So, a subgroup will always contain the identity, in this case denoted 1. If it is non-trivial it will contain other elements too, for instance -1.

So, take the set $\displaystyle \{-1, 1\}$. Firstly, you should note that you have associativity as you inherit this from the group itself. You also have the identity element, this is just 1.

Can you find an inverse for -1? Is this in the set?

What about closure? If you multiply two elements from this set are we still in the set? The only non-trivial product is $\displaystyle (-1)^2$, but this is still easy...
• Jan 5th 2010, 05:09 AM
Quote:

Originally Posted by Swlabr
A subgroup is a set of elements from G which form a group under the operation of G.

So, a subgroup will always contain the identity, in this case denoted 1. If it is non-trivial it will contain other elements too, for instance -1.

So, take the set $\displaystyle \{-1, 1\}$. Firstly, you should note that you have associativity as you inherit this from the group itself. You also have the identity element, this is just 1.

Can you find an inverse for -1? Is this in the set?

What about closure? If you multiply two elements from this set are we still in the set? The only non-trivial product is $\displaystyle (-1)^2$, but this is still easy...

the inverse of -1 is -1 and is in the set

1 times -1 = -1 which is in the set which shows closure

but i thought you said 1 couldnt be in the subset?
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