Help showing that G/H is a group?

[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?

 

[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).

 

[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.)

 

[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?

 

[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?

 

[jbunniii]

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?

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.

Try proceeding this way instead:

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.

 

[millwallcrazy]

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

 

[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.

 

[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?