# Is my approach to this problem correct?

1. Oct 8, 2012

### Jamin2112

1. The problem statement, all variables and given/known data

Suppose H is a normal subgroup G and H has order 2. Show that H is contained in the center of G.

2. Relevant equations

Definitions.

H is a normal subgroup of G if for all h in H and g in G, ghg-1 is in H.

The center of a group G is Z = {z in G | zx = xz for all x in G}.

3. The attempt at a solution

|H| = 2 implies H = {1, h = h-1} for some h. If, in addition, H is a normal subgroup, then for any g in G we have ghg-1 in H = {1, h}. I need to deduce from this that gx = xg for all x in G. Am I right? Is there something missing?

2. Oct 8, 2012

### Zondrina

By "H is a normal subgroup G", do you mean H≤G or H<G? A bit of notational differences here.

Last edited: Oct 8, 2012
3. Oct 8, 2012

### Jamin2112

The definition I posted is straight from the book. But yea, it would mean H is a subset of G.

4. Oct 8, 2012

### Zondrina

Ahh I just wanted to make sure you didnt mean a proper subset.

Instead of 1, you should argue a general identity element e in H. Since H is a subgroup of G, you know that H has an identity element e you can use. Which also has an inverse :).

So, use these to aid you in showing that $e \in H$ and H a subgroup of G, imply that e is also in the center of G. Since the identity element commutes with every other element right?

5. Oct 8, 2012

### I like Serena

Hi Jamin2112!

Seems to me it's a bit sharper.
You have only 1 h in H, so for that h, you should have that $\forall g: gh=hg$.
And since H is normal in G.....

6. Oct 8, 2012

### Jamin2112

So inevitably H = {e, e-1}?

7. Oct 8, 2012

### Dick

e^(-1)=e, yes? H has to contain TWO elements. They can't BOTH be e.

8. Oct 9, 2012

### Jamin2112

Right. e is its own inverse, so to speak.

9. Oct 9, 2012

### Jamin2112

Ok. Here's where I am.

Suppose H has order 2. By the definition of a group, clearly H = {e, h} where e is the identity element in H and h is some other element. Now suppose that H is a normal subgroup of G. Choose g in G. We have ghg-1 and geg-1 both in H. These are distinct elements, the distinct elements in H commute. Hence ghg-1(geg-1) =geg-1(ghg-1).

How wrong am I?

10. Oct 9, 2012

### Dick

You aren't really thinking this through before posting yet another guess at the answer and waiting for someone to correct you. I think that's a pretty bad approach. What is geg^(-1)? Really this time, no guessing allowed.

11. Oct 9, 2012

### Jamin2112

geg^(-1) = e

12. Oct 9, 2012

### Dick

And so? What about ghg^(-1)? Could that be e??

13. Oct 9, 2012

### Jamin2112

No. It would have to be h.

14. Oct 9, 2012

### Dick

You should probably give a reason for that. And once you know that, how does it help you solve your problem?

Last edited: Oct 9, 2012
15. Oct 9, 2012

### Jamin2112

You right multiply by g to obtain gh = hg.

16. Oct 9, 2012

### Dick

Sure, you can write up a proof now, yes?

17. Oct 9, 2012

### Jamin2112

I just wanna be sure of one more thing: How do I know e is G's identity element? I know it's H's identity element and I know it is contained in G, but that doesn't necessarily mean it's G's identity element, no?

18. Oct 9, 2012

### Dick

Odd question. If h is any element of H then hh^(-1)=e, where e is the identity of G. So e MUST be in H. Look at your definition of 'subgroup'.

Last edited: Oct 9, 2012