1. Groups

Hi. Hope someone can help with my problem.

Let G be a group with identity e. If A and B are subgroups of G then, from the Direct Product of Groups Theory, AxB is a subgroup of GxG. Use this result througout.

a) Show that, if A and B are two normal subgroups of G, then AxB is normal subgroup of GxG (Not sure if I am to prove that A and B are two normal subgroups as well as proving AxB is a normal subgroup of GxG).

b) Let H be the subset

H={(g,g):g E G}

Show that H is a subgroup of GxG

Normally I can do these type of questions, it's just the Gx G and AxB bit has confused me a bit.

Thanx

2. Originally Posted by bex23
a) Show that, if A and B are two normal subgroups of G, then AxB is normal subgroup of GxG (Not sure if I am to prove that A and B are two normal subgroups as well as proving AxB is a normal subgroup of GxG).
You need to prove $(g_1,g_2)(a,b)(g_1^{-1},g_2^{-1})\in A\times B$ thus $(g_1ag_1^{-1},g_2bg_2^{-1})\in A\times B$ because $g_1ag_1^{-1}\in A$ and $g_2bg_2^{-1} \in B$.

3. Originally Posted by ThePerfectHacker
You need to prove $(g_1,g_2)(a,b)(g_1^{-1},g_2^{-1})\in A\times B$ thus $(g_1ag_1^{-1},g_2bg_2^{-1})\in A\times B$ because $g_1ag_1^{-1}\in A$ and $g_2bg_2^{-1} \in B$.
I seem to be having a brain dead day, as I can't even remember how to do that

4. Groups

Right, I think I have figured out part a) it's just part b) that is giving me problems. I know that to prove that H is a subgroup of GxG, I need to prove the three subgroup axioms Closure, Identity and Inverse. So far I have only gone so far as Closure and even then I am not sure that this is correct any help will be greatly appreciated.

Closure:

Let (g_1, g_2) and (g_3, g_4) be any two elments of GxG. By the definition of GxG, we know that g_1, g_2, g_3 and g_4 are elemnts of G.

Since G is a group, it is closed under the operation o and we have

(g_1,g_2)o(g_3,g_4)=(g_1 o g_3, g_2 o g_4)
which belongs to H

So the closue axiom holds.

Is this correct and if it is (which I don't think it is) can you help me get started with the other two axioms.

Thanx

Bex

5. It is correct.

For identity element consider $(e,e)$, show it is identity element.
For inverse, note $(g_1,g_2)^{-1} = (g_1^{-1},g_2^{-1})$.

6. Groups

Identity:
To prove the identity axiom we use the identity element (e,e) which belongs to GxG and the arbitrary element
(g_1,g_2) which also belongs to GxG, Thus

(e,e) o (g_1,g_2)= (e o g_1, e o g_2)
=(g_1,g_2) which belongs to H

Inverse:

For this I have two methods (neither which I am very confidant on).

Method1:

To prove the inverse axiom, let (g_1,g_2) be and element of GxG where g_1^-1 and g_2^-1 are the inverses respectively. Proving that (g_1^-1, g_2^-1) is the inverse of (g_1,g_2) in GxG this will prove the inverse axiom (I hope!!).

We have

(g_1,g_2)o(g_1^-1, g_2^-1)=(g_1og_1^-1, g_2og_2^-1)
=(e,e)

Also

(g_1^-1, g_2^-1)o(g_1,g_2)=(e,e)

Hence, (g_1^-1,g_2^-1) is the inverse of (g_1,g_2)

Method 2:

Let (g_1,g_2) be an element of GxG then the inverse is
(g_1^-1,g_2^-1), which belongs to H as it is of the form (g,g) for g E G.

I think method 2 is the more wrong of the two but put in in here just in case.

Have I gone completly off the mark here. I was surprised when you said that the first part was right, and so now the rest will probably be all wrong. Lol!! Can you just tell me if I have gone in the right direction, and if not point me in the right one.

Thanx

Bex