# Showing an operator is well-defined

1. Nov 21, 2014

### Bashyboy

1. The problem statement, all variables and given/known data
Let $H$ be a normal subgroup of a group $(G, \star)$, and define $G/H$ as that set which contains all of the left cosets of $H$ in $G$. Define the binary operator $\hat{\star}$ acting on the elements of $G/H$ as $g H \hat{\star} g' K = (g \star g') H$.

2. Relevant equations

3. The attempt at a solution
I am having difficulty demonstrating that $\hat{\star}$ is well-defined. I understand that it is two cosets to be equal. So, for instance, we could have

$g_1 H = g_2 H$

and

$g_3 H = g_4 H$

So, to show that the operator is well-defined, I would have to show that

$g_1 H \hat{\star} g_3 H = g_2 H \hat{\star} g_4 H$

is true. However, I am having difficulty with this. If I understand correctly, $g_1 H = g_2 H$ does not necessarily imply that $g_1 = g_2$.

2. Nov 21, 2014

### haruspex

$g H \hat{\star} g' H = (g \star g') H$ presumably.
Yes, but write out what that means according to the definition of $\hat{\star}$. Then consider what $(g_1 \star g_3) H$ is as a set.

3. Nov 21, 2014

### Bashyboy

So, let me see if I understand correctly. $g_1 H~ \hat{\star} ~g_3 H = (g_1 \star g_3) H$ is simply some set, as well as $g_2 H~ \hat{\star}~ g_4 H = (g_2 \star g_4) H$; and we are trying to establish the equality of the sets $(g_1 \star g_3) H$ and $(g_2 \star g_4) H$. Would we do this by considering an arbitrary element in each set, and then showing that they are equivalent?

4. Nov 21, 2014

### haruspex

Yes.
No. It will not in general be true that $g_1 \star g_3 \star g = g_2 \star g_4 \star g$.
Write out what $g_3 H$ means in set-theoretic notation (like, {some expression involving h : h element of H}), then extend that to what $(g_1 \star g_3) H$ means.

5. Nov 21, 2014

### Dick

Use that H is NORMAL. That's very important.

6. Nov 22, 2014

### Bashyboy

Well, the set $g_3H$ is $\{g_3 h | h \in H\}$, and $(g_1 \star g_3) H$ would be $\{g_1 g_3 h | h \in H \}$. Because $H$ is normal, that means every $g \in G$ and every $h \in H$, there exists and $h_1 \in H$ such that $h$ can be written as $h = g h_1 g^{-1}$.

Does this mean that there are many representations of $h$? Does this mean that I should be able to write $h$ in a similar manner as above for any $g$? For instance, I could write $h = g_1 h_1 g_1^{-1}$, or $h = g_2 h_1 g_2^{-1}$?

7. Nov 22, 2014

### Dick

Not with the same $h_1$. I usually think of the definition of $H$ being normal as $gH=Hg$ for any element $g$. Do you see how H being normal implies that if $g_1 H=g_2 H$ then there are elements of $H$, $h$ and $h'$ such that $g_1=g_2 h$ and $g_1=h' g_2$?

8. Nov 22, 2014

### Bashyboy

I realize that if $H$ is normal, then, as you said, $gH = Hg$ $\forall g \in G$. In particular, $g_2 H = H g_2$. Returning to the assumption that $g_1 H = g_2 H$, this can be written as $g_1 H = Hg_2$.

So, every element in $g_1 H$ will also be in $g_2 H$ and $Hg_2$. So, because $g_1 \in g_1 H$, then it must be true that there exist $h_1 , h_2 \in H$ such that $g_1 = g_2 h_1$ and $g_1 = h_2 g_2$, as you have written. But I am having difficulty connecting that with what we have already covered.

Last edited: Nov 22, 2014
9. Nov 22, 2014

### Dick

You want to show that if $g_1 H=g_2 H$ and $g_3 H=g_4 H$ then $g_1 g_3 H=g_2 g_4 H$. Try to convert one side into the other.

Last edited: Nov 22, 2014
10. Nov 23, 2014

### Bashyboy

I am sorry, but this problem is giving me great difficulty. We know that $g_1 = g_2 h_1$ and $g_1 = g_2 h_2$, allowing me to make the substitutions:

$(g_1 \star g_3 ) H = (g_2 h_1 \star g_3) H$

or

$(g_1 \star g_3 ) H = (h_2 g_2 \star g_3)H$

But I do not see how either would be helpful, unless I could somehow eliminate $h_1$ and $h_2$.

11. Nov 23, 2014

### Dick

What is $h H$ where $h$ is an element of $H$?

12. Nov 23, 2014

### Bashyboy

Wouldn't it be that $hH = H$? No, because if $h \ne e_H$, then $hH$ would not contain the identity element.

13. Nov 23, 2014

### Dick

Think again. $h^{-1}$ is in $H$.

14. Nov 23, 2014

### Bashyboy

Oh, of course! So, then how can I use this fact to simplify $(g_2 h_1 \star g_3)H$? Would I write $(g_2 h_1 \star g_3)H = g_2 H ~ \hat{\star} ~h_1 H~ \hat{ \star}~ g_3 H$. And then I can use the fact that $h_1 H = H = e_H H$. Finally, I would do a similar substitution for $g_3$. Does sound correct?

15. Nov 23, 2014

### Dick

It's a little awkward. How about $g_2 h_1 g_3 H=g_2 h_1 H g_3=g_2 H g_3=g_2 g_3 H$? Can you justify those steps?

16. Nov 23, 2014

### Bashyboy

Yes, I can. You seem to use the fact that $g_3 H = H g_3$ twice, and the fact that $h_1 H = H$ in between using the aforementioned fact. Is that right?

Last edited: Nov 23, 2014
17. Nov 23, 2014

### Dick

Exactly.

18. Dec 1, 2014

### Bashyboy

Hold on. I don't understand why we have to use the fact that $H$ is normal. We know that $g_1 \in g_1 H$, and because $g_1 H = g_2$, we know that $g_1 = g_2 h_1$; similarly, $g_3 = g_4h_2$

$g_1 H ~ \hat{\star} ~ g_3 H = g_2 h_1 H ~ \hat{\star} ~ g_4h_2 H$

Because $hH = H$ $\forall h \in H$, we get

$g_1 H~ \hat{\star}~ g_3 H = g_2 H~ \hat{\star}~ g_4 H$

Thus, the different representations get mapped to the same thing.

19. Dec 1, 2014

### Dick

Indeed. But you also need that $g_1 H g_3 H=g_1 g_3 HH=g_1 g_3 H$. That shows that the product of two cosets is a coset. If $H$ isn't normal that's not necessarily true.

20. Dec 1, 2014

### Bashyboy

$g_1 H g_3 H=g_1 g_3 HH=g_1 g_3 H$

I don't understand what your wrote in the middle. In the problem statement they define $\hat{\star}$ as $gH \hat{\star} g' H = (g \star g')H = gg'H$