Subgroup

**jzellt** · October 10th 2010, 07:32 PM

Let G be an abelian group.

If H = { x e G : x = x^-1 }, that is, H consists of all the elements of G which are their own inverses. Prove that H is a subgroup of G.

**Jhevon** · October 10th 2010, 07:38 PM

Originally Posted by jzellt

Let G be an abelian group.

If H = { x e G : x = x^-1 }, that is, H consists of all the elements of G which are their own inverses. Prove that H is a subgroup of G.

you need to show three things: H is nonempty (some texts require that you check that the identity be in the subgroup...which has to happen anyway, and still fulfills the nonempty condition), and closed under products and closed under inverses. 2 are obvious. where did you begin? where are you stuck?

**jzellt** · October 10th 2010, 08:13 PM

I honestly don't know how to show those things. Could you show me? Thanks.

**Jhevon** · October 10th 2010, 08:18 PM

Originally Posted by jzellt

I honestly don't know how to show those things. Could you show me? Thanks.

Come on, how about you try.

Try the first one: the identity, e, is in G. How would you show this? What properties of the identity would ensure that it is in G?

**jzellt** · October 10th 2010, 08:30 PM

Im sorry. I missed the lecture when she talked about this. Im just trying to see how this one is done so I can do the others.

Do i just say that e is in G since e = e^-1 ??

**Jhevon** · October 10th 2010, 08:44 PM

Originally Posted by jzellt

Im sorry. I missed the lecture when she talked about this. Im just trying to see how this one is done so I can do the others.

Do i just say that e is in G since e = e^-1 ??

Yes! the identity is its own inverse. so it is in H by definition. (we want it to be in H, not G. i made a mistake in my last post. H is what we want to show is a subgroup)

the next easiest condition to me is closure under inverses. do you know what that means? try to show it

**jzellt** · October 10th 2010, 08:50 PM

I think that means that every element in H has an inverse, but I'm not 100% sure... Don't know how to show this though...

**Jhevon** · October 10th 2010, 08:53 PM

Originally Posted by jzellt

I think that means that every element in H has an inverse, but I'm not 100% sure... Don't know how to show this though...

it means, that if $\displaystyle g \in G$ , then $\displaystyle g^{-1} \in G$ . So yes, for each element in the set, its inverse will be in the set. if this happens, we say the set is "closed under inverses"

So now, is this true for $\displaystyle H$ ? If $\displaystyle x \in H$ , is it true that $\displaystyle x^{-1} \in H$ ?

**jzellt** · October 10th 2010, 09:05 PM

I would say yes... x^-1 e H. Since H consists of all elements of G which their own inverses, then if x e H, then x^-1 e H. This probably isn't a sufficient proof or is completely wrong... Can you show how to show this

**Jhevon** · October 10th 2010, 09:20 PM

Originally Posted by jzellt

I would say yes... x^-1 e H. Since H consists of all elements of G which their own inverses, then if x e H, then x^-1 e H. This probably isn't a sufficient proof or is completely wrong... Can you show how to show this

geez, have some self confidence, man! yes, that's right. if x is in H, then so will its inverse be in H, since x IS its own inverse.

now for the last condition. This one is harder, i will do it. First note, that $\displaystyle x = x^{-1}$ is the same as saying $\displaystyle x^2 = xx = e$ . also note, that $\displaystyle G$ is Abelian if the operation on it is commutative. That is, $\displaystyle xy = yx$ for all $\displaystyle x,y \in G$ . Now on to the proof.

We show that $\displaystyle H$ is closed under products (that is, if $\displaystyle x \in H \text{ and }y \in H$ , then $\displaystyle xy \in H$ ). We need to show that $\displaystyle x,y \in H \implies xy \in H$ , we will do this by showing that $\displaystyle xy = (xy)^{-1}$ , provided $\displaystyle x$ and $\displaystyle y$ are in $\displaystyle H$ .

Assume $\displaystyle x,y \in H$ . Since $\displaystyle x,y \in G$ , and $\displaystyle G$ is Abelian, we have $\displaystyle xy = yx$ .

$\displaystyle \Rightarrow xyx = yxx$ .........multiplied both sides by x

$\displaystyle \Rightarrow xyx = y$ ............since $\displaystyle xx = e$ , because $\displaystyle x \in H$

$\displaystyle \Rightarrow xyxy = yy$ ..........muliplied both sides by y

$\displaystyle \Rightarrow xyxy = e$ ..........since $\displaystyle yy = e$ , because $\displaystyle y \in H$ .

$\displaystyle \Rightarrow (xy)^2 = e$ , as desired.

This completes the proof.

**jzellt** · October 10th 2010, 09:24 PM

Thanks a lot for your help!!

So, in general, anytime I'm trying to show that a set is a subgroup of another, there's only 3 things to show no matter what?

1. e is an element
2. closed under inverses
3. closed under products

**Jhevon** · October 10th 2010, 09:37 PM

Originally Posted by jzellt

Thanks a lot for your help!!

So, in general, anytime I'm trying to show that a set is a subgroup of another, there's only 3 things to show no matter what?

1. e is an element
2. closed under inverses
3. closed under products

yes. and to show something is a group, there are 4 things to show. do you know these?

**jzellt** · October 10th 2010, 09:46 PM

Group is just 3 things right?
Associativity, identity, and inverse

Abelian Group has 4 - Commutativity and the above 3. Correct?

1 last thing: Can give you give me a pointer on how to show that e is an element of H if H = {x e G : x^n = e} (n is a fixed integer)

**Jhevon** · October 10th 2010, 10:01 PM

Originally Posted by jzellt

Group is just 3 things right?
Associativity, identity, and inverse

Abelian Group has 4 - Commutativity and the above 3. Correct?

1 last thing: Can give you give me a pointer on how to show that e is an element of H if H = {x e G : x^n = e} (n is a fixed integer)

for a group, you have to show closure also. and for identity, you have to show it exists. this is different than with subgroups, where you know the identity exists, you just want to show it is in the subgroup.

no pointer really, take e^n = e is true for any integer n

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