# Help showing that G/H is a group?

1. May 14, 2010

### millwallcrazy

Hi guys

I was just wondering whether anyone could help me with Group Theory

I am trying to prove that G/H is a group (where G is a group and H is a normal subgroup of G)

I know that i need to go through the 4 properties, Identity, inverse, Associativity and Closure but i'm not sure where to start. I'm also not sure how to properly define the composition Law in G/H

Is anybody able to help?

2. May 14, 2010

### Tedjn

The elements of G/H are cosets of the form aH, where a is in G. If aH and bH are in G/H, the composition law is given by (aH)(bH) = (ab)H. In addition to whatever group properties, you must also show that this law of composition is well-defined. That is, if aH = a'H and bH = b'H, then (aH)(bH) = (a'H)(b'H).

3. May 14, 2010

### jbunniii

First start by defining the group operation.

The elements of G/H are the cosets of H. (Let's say left cosets for definiteness. It actually doesn't matter because left cosets and right cosets are the same thing for normal subgroups, but I don't know if you have proved that yet.)

Thus $$G/H = \{gH : g \in G\}$$.

I claim that the following is a well-defined operation on G/H:

$$(aH)(bH) = (ab)H$$

However, before you can proceed, you need to verify that this is well-defined. Why is this an issue? It's because the coset $aH$ has multiple elements (assuming H is not trivial), and $a$ is only one of the elements. Therefore you must show that if $a_1 H = a_2 H$ and $b_1 H = b_2 H$, then $(a_1 b_1)H = (a_2 b_2)H$.

Once you have done that, you can proceed with verifying that the group axioms hold. (Hint: H is the identity element.)

4. May 14, 2010

### millwallcrazy

Thanks for the help, i was just wondering how i go about showing that it was well defined. I tried to use the facts that were given

I didn't know what to do once (aH)(bH) = (ab)H....is the next step to say that this is ab'H?

5. May 14, 2010

### millwallcrazy

I have now come to this conclusion:

i need to show that (ab)H = (a'b')H

First step: (ab)H = aHbH = a'Hb'H = (a'b')H

Is that how to show its well defined or is there anything that i wasnt allowed to do?

6. May 14, 2010

### jbunniii

No, if the goal is to show that an operation is well-defined, you can't use that operation (thereby implicitly assuming it's well-defined) as part of the proof. Also, you didn't use the fact that H is a normal subgroup of G, which should be a red flag that something is wrong.

We assume that aH = a'H and bH = b'H.

Then

abH = ab'H

Now, *if* it were true that b'H = Hb', then we could continue this way:

abH = ab'H = aHb' = a'Hb' = a'b'H

and we would be done.

So all we need to do is show that b'H = Hb'. How do we do that? Hint: we still haven't used the fact that H is normal.

7. May 14, 2010

### millwallcrazy

For a group to be normal subgroup of another doesnt it mean gHg-1 belongs to H? I don't see how this would help?

8. May 14, 2010

### jbunniii

OK, so your definition is that H is normal in G if

$$gHg^{-1} \subset H$$

for every $g \in G$, right?

I claim that something stronger is true:

$$gHg^{-1} = H$$

for every $g \in G$. Assuming my claim is true, then I can simply multiply both sides on the right by $g$ to get

$$gH = Hg$$

(Here I used $g^{-1}g = 1$.)

Now substitute $b'$ for $g$ and you get the desired

$$b'H = Hb'$$

So now you just need to prove my claim. I suggest starting with

$$gHg^{-1} \subset H$$

and think about how you can morph that into

$$H \subset g^{-1}Hg$$

for all $g \in G$.

9. May 15, 2010

### millwallcrazy

Can we just multiply both sides on the left by g-1 and the right by g so that it now becomes the required result? i.e. H is a subset of g-1Hg?