Showing something is a subgroup of the direct product

[kathrynag]

Homework Statement

Let G1 and G2 be groups, with subgroups H1 and H2 respectively. Show
that {(x1,x2) such that x1 is in H1, x2 is in H2} is a subgroup of the
direct product G1 x G2

Homework Equations

The Attempt at a Solution

let G1, G2 be groups with H1, H2 subgroups.
Let (x1,x2) such that x1 is in H1 and x2 is in H2
By definition of a direct product, we can say H1 x H2=(x1,x2)
We want G1 x G2= (x1,x2). Not sure how to get there.

 

[micromass]

What do you need to prove for something to be a subgroup?

 

[kathrynag]

all group properties hold under the operation defined

 

[micromass]

Yes, but there's a shorter way.

Let G be group and $$H\subseteq G$$, then H is a subgroup of G if and only if
- $$1\in H$$ where 1 is the neutral element of G
- $$a,b\in H~\Rightarrow~a.b\in H$$
- $$a\in H~\Rightarrow a^{-1}\in H$$

Did you see that?

These three properties hold for H1 and H2. Can you show that they hold for H1xH2?

 

[kathrynag]

So we have H1 x H2 = (x1,x2)
1 is in H1 x H2 since we can have (1,1) since 1 is in H1 and H2
(x1,x2) , (y1,y2) is in H1 x H2
Then (x1y1,x2y2) is H1 x H2
Not sure if I follow this step
(x1,x2) is in H1 x H2.
Since x1,x2 are in H1 and H2 respectively, inverses exist
(x1x^-1,x2,x2x2^-1)=(e,e)=(ex^-1,ex^-1)=(x1^-1,x2^-1)

 

[micromass]

Everythings good, except the last step.

Take an element (x1,x2) in H1 x H2. We need to find the inverse of these elements. Since H1 and H2 are subgroups, we know that x1^-1 and x2^-1 are in H1 and H2 respectively. So the element (x1^-1,x2^-1) is in H1xH2. Can you prove that this element is the inverse of (x1,x2)??

 

[kathrynag]

ok by inverse a^-1*a=e
So if (x1^-1,x2^-1) is an inverse, we must have (x1^-1,x2^-1)(x1,x2)=(e,e)
(x1^-1x1,x2^-1x2)=(e,e) since x1 and x1^-1 are inverses of H1, so x1*x1^-1=x1^-1x1=e. Similar argument for x2.

 

[micromass]

Alright, I think you've got it.