# coset question

• Jul 23rd 2009, 03:15 AM
sung
coset question
he guys,

I make an error in my reasoning but I can't see what i do wrong:

G is a group and H is a subgroup. Now we have to left cosets, I have to show that they are also the right cosets.
Now G has to be the union of the sets 1*H and g*H with g not in H.
now I take g1 also not in H.
Then I show that g1*h1=g*h (for h1,h in H)
but then h=(g^-1)*g1*h1. But then (g^-1)*g1 is in H en I get the first coset 1*H. So is (g^-1)*g1 not in H? What is wrong with this?

Thanks for the help
• Jul 23rd 2009, 04:35 AM
Swlabr
Quote:

Originally Posted by sung
he guys,

I make an error in my reasoning but I can't see what i do wrong:

G is a group and H is a subgroup. Now we have to left cosets, I have to show that they are also the right cosets.
Now G has to be the union of the sets 1*H and g*H with g not in H.
now I take g1 also not in H.
Then I show that g1*h1=g*h (for h1,h in H)
but then h=(g^-1)*g1*h1. But then (g^-1)*g1 is in H en I get the first coset 1*H. So is (g^-1)*g1 not in H? What is wrong with this?

Thanks for the help

Cosets are just sets, they are never (with the single exception of $\displaystyle 1*H$) groups under the multiplication from $\displaystyle G$. For instance, there is no identity in the coset $\displaystyle g*H$, $\displaystyle g \notin H$. Thus, there is no reason for $\displaystyle g^{-1}*g1$ to not be in $\displaystyle H$. In fact, two elements are in the same coset if and only if $\displaystyle g_1^{-1}g_2 \in H$. You have just proved one way of this!

This example is a very special example. In the general case, we would use normality, but here we are not allowed to as H is not stated as being normal. It turns out, however, that H must be normal (in fact, this is what you are proving. If you prove that $\displaystyle gH=Hg$ for all $\displaystyle g \in G$, as you are doing here, then $\displaystyle gHg^{-1}=H$, and so $\displaystyle H$ is normal).

Clearly, $\displaystyle 1*H=H*1$. We thus only require to show that $\displaystyle Hg=gH, g \notin H$. To do this, we must prove two things:

1) Prove that $\displaystyle Hg \cap H = \emptyset$ and $\displaystyle gH \cap H = \emptyset$.

2) Prove that $\displaystyle |Hg|=|gH|$ (alternatively, prove that there can only exist two right cosets).

Can you see why this gives us the result?

The proofs of these two results are not hard, and this site exists merely to help you understand (not to provide answers), and so I will omit the proofs for these two results. They are, however, not too hard (although the second one is a bit tricky).
• Jul 23rd 2009, 08:58 AM
HallsofIvy
One serious problem you have is that the statement you are trying to prove is NOT true. In general, it not true that "left cosets" of a subgroup are also right cosets of that same subgroup. When they are, that is a very important special case- we say that H is a normal subgroup of G if and only if the left and right cosets of H are the same.

(Of course, if a group is Abelian, them "left" and "right" are the same- left and right cosets are the same. All groups of order 5 or less are Abelian so you must go up to order to 6 to have a chance at an example. But, if I remember correctly, since 6= 2(3), the product of two primes, a group of order 6 also has all subgroups normal. Of course 7 is itself prime and therefore Abelian (all groups of prime order are cyclic groups so you really, I think, need to go to order 8 to find a group in which the left cosets are different from the right cosets.)
• Jul 23rd 2009, 09:19 AM
Swlabr
Quote:

Originally Posted by HallsofIvy
One serious problem you have is that the statement you are trying to prove is NOT true. In general, it not true that "left cosets" of a subgroup are also right cosets of that same subgroup. When they are, that is a very important special case- we say that H is a normal subgroup of G if and only if the left and right cosets of H are the same.

Read between the lines - we are dealing with a subgroup of index two,

Quote:

Originally Posted by sung
Now G has to be the union of the sets 1*H and g*H with g not in H.

$\displaystyle G$ is the union of two cosets, $\displaystyle 1H$ and $\displaystyle gH$, and thus $\displaystyle H$ is of index 2, $\displaystyle |G:H|=2$. Every subgroup of index two is normal, as I go some way to proving in my earlier post.
• Jul 23rd 2009, 09:29 AM
Swlabr
Quote:

Originally Posted by HallsofIvy
(Of course, if a group is Abelian, them "left" and "right" are the same- all cosets. All groups of order 5 or less are Abelian so you must go up to order to 6 to have a chance at an example. But, if I remember correctly, since 6= 2(3), the product of two primes, a group of order 6 also has all subgroups normal. Of course 7 is itself prime and therefore Abelian (all groups of prime order are cyclic groups so you really, I think, need to go to order 8 to find a group in which the left cosets are different from the right cosets.)

In $\displaystyle D_8$, the dihedral group of order 8, the flip element (the "standard" generating element of order two) generates a non-normal subgroup.

$\displaystyle D_8=<\alpha, \beta: \alpha^4=\beta^2=1, \beta \alpha \beta = \alpha^{-1}>$

$\displaystyle \alpha^{-1} \beta \alpha = \beta \alpha \beta \beta \alpha = \beta \alpha^2$.

Clearly $\displaystyle \beta \alpha^2 \neq \beta$ as otherwise $\displaystyle \alpha^2=1$, a contradiction. Further, $\displaystyle \beta \alpha^2 \neq 1$ as else $\displaystyle \alpha^2 = \beta$ which gives us that $\displaystyle \alpha^{-1} = \beta \alpha \beta = \alpha^5=\alpha$ and so $\displaystyle o(\alpha)=2$, again a contradiction. Thus, $\displaystyle \alpha^{-1} \beta \alpha \notin <\beta>$.

Thus, $\displaystyle D_8$ has a non-normal subgroup, $\displaystyle <\beta>$...
• Jul 23rd 2009, 02:18 PM
alunw
Unless I am very much mistaken the dihedral group of order 6 has non-normal subgroups and distinct left and right cosets. Consider the group using the presentation a^2 = b^3 = (ab)^2 = 1 and denote b^-1 by B.

Then the subgroup {1,a} is not normal since b*a*B = a*B*B = a*b and we have Hb = {b,ab} and bH = {b,aB}