Abstract Algebra: Proving whether H is a subgroup.

[The_Iceflash]

Homework Statement

Let R = {all real numbers}. Then <R,+> is a group. (+ is regular addition)
Let H = {a|a $$\epsilon$$ R and a² is rational}.

Is H closed with respect to the operation?
Is H closed with respect to the inverse?
Is H a subgroup of G?

Homework Equations

N/A

The Attempt at a Solution

For the inverse:

From my notes I know that If it is closed in respect to the inverse that I'll have to prove:
If x$$\epsilon$$ H, then x inverse $$\epsilon$$ H. If it's not I'll have to show an element b $$\epsilon$$ H, but b^-1 is not an element of H.
I know for the inverse that a^.1 = -a for addition. It's looking like it's yes since a and -a is and element of R but I have to prove that somehow and I'm not sure how.

For the operation:

My notes show for operation that H is closed in respect to the operation if a * b $$\epsilon$$ H for any a,b $$\epsilon$$ H.

Knowing that I wrote a+b $$\epsilon$$ H for any a,b $$\epsilon$$ H. Now I'm pretty sure that this is also yes but I'm going to have to prove if x,y $$\epsilon$$ H, then x +y $$\epsilon$$ H and I'm not sure how to actually prove it.

I know that if both of these is 'yes' then it is indeed a subgroup.

Help is greatly appreciated.

 

[CompuChip]

First a general remark: instead of showing that a non-empty subset H of a group G satisfies all the axioms, it is sufficient to show that $x y^{-1} \in H$ for all x and y in H.

For the first question, how about $a = \sqrt{2}, b = \sqrt{3}$ ?

For the second one, just use the definition. Suppose a is in H, then a² is rational. Is (-a) in H then, i.e. is (-a)² rational?

 

[The_Iceflash]

First a general remark: instead of showing that a non-empty subset H of a group G satisfies all the axioms, it is sufficient to show that $x y^{-1} \in H$ for all x and y in H.

Ok good. First to clarify notation, is it the inverse of x time y or is it x times the inverse of y?
How would I go about showing that?

For the first question, how about $a = \sqrt{2}, b = \sqrt{3}$ ?

So this is a counterexample that will prove it isn't an operation?

For the second one, just use the definition. Suppose a is in H, then a² is rational. Is (-a) in H then, i.e. is (-a)² rational?

Oh I understand. How would you actually recommend proving that though?

 

[CompuChip]

Sorry, it means x times the inverse of y.
It's a good exercise in group theory to prove that statement (or look up the proof and follow it closely).About your last question... take another look. We know that a² is rational. How about (-a)² ?

 

[The_Iceflash]

If I did this correctly I was wrong for both of them.

For the operation I did:

$$\sqrt{2}$$, $$\sqrt{3}$$ $$\in$$ H but $$\sqrt{2}$$ + $$\sqrt{3}$$ is not in H.

Reason: $$\sqrt{2}$$ + $$\sqrt{3}$$ is a real number but , $$\left(\sqrt{2}+\sqrt{3}\right)^{2}$$ is not rational.
Therefore H is not closed with respect to the operation.


For the inverse I did:

H is closed with respect to the inverse:

Suppose a is in H then a$$^{2}$$ is rational. If -a is in H, then (-a)$$^{2}$$ is rational.

Since 0 is the identity, a + (-a) = 0
So if 'a' is $$\in$$ H, a$$^{-1}$$ $$\in$$ H

and since H isn't closed with respect to both the inverse and the operation it is not a subgroup of G.

How's that?

 

[CompuChip]

If I did this correctly I was wrong for both of them.

For the operation I did:

$$\sqrt{2}$$, $$\sqrt{3}$$ $$\in$$ H but $$\sqrt{2}$$ + $$\sqrt{3}$$ is not in H.

Reason: $$\sqrt{2}$$ + $$\sqrt{3}$$ is a real number but , $$\left(\sqrt{2}+\sqrt{3}\right)^{2}$$ is not rational.
Therefore H is not closed with respect to the operation.

Agreed, although it would be nice if you showed this explicitly. For example, if you'd take $a = b = \sqrt{2}$ then (a + b)² would be rational...

For the inverse I did:

H is closed with respect to the inverse:

Suppose a is in H then a$$^{2}$$ is rational. If -a is in H, then (-a)$$^{2}$$ is rational.

Since 0 is the identity, a + (-a) = 0
So if 'a' is $$\in$$ H, a$$^{-1}$$ $$\in$$ H

This is still a bit ... fuzzy. All your statements are true, but where do you prove that (-a) is in H? I.e. where can I read that (-a)² is rational?

 

[The_Iceflash]

Agreed, although it would be nice if you showed this explicitly. For example, if you'd take $a = b = \sqrt{2}$ then (a + b)² would be rational...

Oh. ok.

is still a bit ... fuzzy. All your statements are true, but where do you prove that (-a) is in H? I.e. where can I read that (-a)² is rational?

Oh that's right.

I should have said that:
For <R,+>: a+0=a thus making 0 the identity.
then a+x=0 for determining the inverse.
thus -a would be the the inverse for the operation.

I'm not sure where to go from there.

 

[vela]

I should have said that:
For <R,+>: a+0=a thus making 0 the identity.
then a+x=0 for determining the inverse.
thus -a would be the the inverse for the operation.

I'm not sure where to go from there.

You're getting sidetracked proving what you already know. You're given that (R,+) is a group. That means you already know that -a is the additive inverse and 0 is the additive identity. The only question you need to answer is: if a is in H, is -a also in H?

 

[The_Iceflash]

Thanks.

 

[The_Iceflash]

You're getting sidetracked proving what you already know. You're given that (R,+) is a group. That means you already know that -a is the additive inverse and 0 is the additive identity. The only question you need to answer is: if a is in H, is -a also in H?

Yes. That's actually where I'm stuck right now. I need to show why -a is in H.

I have that

Since for <R,+> e = 0, then a +a^-1 = 0 and therefore a^-1= -a

What I need to show now is that -a is in H and I don't know how to begin doing that.

For the second part:

For the operation, I need to show a counter example to show that it is not closed in respect to the operation.

I have to show a counter example that shows that it is not rational and therefore not in H. I had $$\sqrt{2}$$ and $$\sqrt{3}$$ being added together and then squared to show it but it doesn't show why. Any ideas on what I can do?

Both of these concerns are open to anyone who can help.

 

[vela]

Yes. That's actually where I'm stuck right now. I need to show why -a is in H.

I have that

Since for <R,+> e = 0, then a +a^-1 = 0 and therefore a^-1= -a

What I need to show now is that -a is in H and I don't know how to begin doing that.

What's the criterion for whether any element is in H? Does -a satisfy that criterion?

For the second part:

For the operation, I need to show a counter example to show that it is not closed in respect to the operation.

I have to show a counter example that shows that it is not rational and therefore not in H. I had $$\sqrt{2}$$ and $$\sqrt{3}$$ being added together and then squared to show it but it doesn't show why. Any ideas on what I can do?

Both of these concerns are open to anyone who can help.

What do you mean it doesn't show why?

 

[CompuChip]

Hey The_Iceflash,

When I read your replies, I begin to suspect that the problem lies in your understanding of the definition. Can you tell me whether the following numbers are or are not elements of H?

$$2, \sqrt{2}, \sqrt{\sqrt{2}}, \pi^3, \frac{4}{3}, \frac{\sqrt{3}}{2}, \left( \frac{1}{7} \right)^{1/3}$$