Is my approach to this problem correct?

[Jamin2112]

Homework Statement

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

Homework 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}.

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?

 

[STEMucator]

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

 

[Jamin2112]

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

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

 

[STEMucator]

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

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?

 

[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...

 

[Jamin2112]

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

 

[Dick]

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

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

 

[Jamin2112]

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

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

 

[Jamin2112]

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...

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?

 

[Dick]

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?

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.

 

[Jamin2112]

What is geg^(-1)? Really this time, no guessing allowed.

geg^(-1) = e

 

[Dick]

geg^(-1) = e

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

 

[Jamin2112]

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

No. It would have to be h.

 

[Dick]

No. It would have to be h.

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

 

[Jamin2112]

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

You right multiply by g to obtain gh = hg.

 

[Dick]

You right multiply by g to obtain gh = hg.

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

 

[Jamin2112]

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

I just want to 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?

 

[Dick]

I just want to 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?

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'.