1. ## Subgroup

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.

2. 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?

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

4. 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?

5. 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 ??

6. 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

7. 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...

8. 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 $\displaystyle \displaystyle g \in G$, then $\displaystyle \displaystyle g^{-1} \in G$. 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 $\displaystyle \displaystyle H$? If $\displaystyle \displaystyle x \in H$, is it true that $\displaystyle \displaystyle x^{-1} \in H$ ?

9. 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

10. 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 $\displaystyle \displaystyle x = x^{-1}$ is the same as saying $\displaystyle \displaystyle x^2 = xx = e$. also note, that $\displaystyle \displaystyle G$ is Abelian if the operation on it is commutative. That is, $\displaystyle \displaystyle xy = yx$ for all $\displaystyle \displaystyle x,y \in G$. Now on to the proof.

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

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

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

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

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

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

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

This completes the proof.

11. 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

12. 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?

13. 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)

14. 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