# Proof: H is a subgroup of G

1. Mar 17, 2012

### jr16

1. The problem statement, all variables and given/known data
Let <G, *> be an Abelian group with the identity element, e. Let H = {g ε G| g2 = e}. That is, H is the set of all members of G whose squares are the identity.

(i) Prove that H is a subgroup of G.
(ii) Was being Abelian a necessary condition?

2. Relevant equations
For any subset H of G there are four conditions it must satisfy in order to be classified as a subgroup of G:
1. H is closed under *
2. * is associative
3. The identity, e, of G is also in H
4. For each a in H, the inverse of a, a-1, is also in H

3. The attempt at a solution
(i) In an attempt to solve this problem, I tried to prove each of the above four conditions:
1. g2*g2
g2 + 2
g4
(g2)2
Therefore, H is closed under *

2. (g*g)*g = g*(g*g)
e * g = g * e
g = g
Thus, * is associative

3. e*e = e
e = e
Yes, e is in H

4. g*g = e
(g-1)*g*g = (g-1)*e
e * g = (g-1)
g = (g-1)
Therefore, (g-1) is in H

Thus, H is a subgroup of G.
(ii) No, G being Abelian was not a necessary condition (???)

I would really appreciate it if someone could look over my above work. I am not confident in the answer I provided, especially since I did not use the Abelian condition. Thank you for any guidance!

2. Mar 17, 2012

### A. Bahat

Careful here: you are confusing H={g ε G: g^2=e} with a different set, {g^2: g ε G}.

3. Mar 17, 2012

### jr16

g2*g2 = e
g2 + 2 = e
g4 = e
(g2)2 = e
Therefore, H is closed under *

does this fix the issue?

4. Mar 17, 2012

### HallsofIvy

Staff Emeritus
All you have done is show that g times itself is in H. That does NOT show that H is closed under multiplication.

Suppose a and b are in H. Then $a^2= 1$ and $b^2= 1$. So what can you say about $(ab)^2$?

By the way, as the conditions for a subgroup you give
You do NOT need to prove (2), that the operation is associative. Since that is the operation "inherited" from G you alredy know that the operation is associative from the fact that G is a group.

5. Mar 17, 2012

### jr16

Thank you, all of this input is very helpful!

In my original post I should have made it more clear that * in this case does not represent multiplication but rather any binary operation on G. Therefore, sn = s*s*s...n times and sa * sb = s(a+b)

Let me try again:
Let a, b be elements of the set H such that a2=e and b2=e

a*b = e
a * b = a * a replacing e with a*a
b = a by the left cancellation law
?

Or should I begin with a2 and b2?
so that
a2 * b2 = e
e * e = e
e = e
?

6. Mar 18, 2012

### A. Bahat

Not quite. You have to show that if a,b are in H, then a*b is in H. You can't assume that a*b=e. Hint: consider (a*b)*(a*b) and use the fact that * is commutative.

7. Mar 18, 2012

### jgens

Let G = D6 and notice that s2 = e and that (sr)2=(sr)(r-1s) = e. But s(sr) = r and r2 ≠ e.

8. Mar 18, 2012

### jr16

Thank you all again for your great responses!

I had a small epiphane over breakfast this morning haha
It involves something that A. Bahat just brought up, so maybe I finally have it right this time?

Prove: a * b = e
(a * b)2 = e2 by squaring both sides
(a * b) * (a * b) = e * e
a * (b * a) * b = e by the associative property
a * (a * b) * a = e by the commutative property (thus, Abelian condition is necessary)
(a * a) * (b * b) = e by the associative property
(e) * (e) = e
e = e
Therefore, H is closed under *
?

9. Mar 18, 2012

### jgens

You are not trying to prove that $a,b \in H$ implies $ab = e$. You are trying to prove that if $a,b \in H$, then $(ab)^2 = e$. You can do this in essentially the same way as your work above indicates.

You should also note that just because you use the fact that $G$ is abelian does not mean that it is necessary for the proof. It suggests that it is necessary, but you need to provide an example where the result fails if $G$ is non-abelian in order to show that $G$ abelian was necessary.

10. Mar 18, 2012

### jr16

ah OK, let's give this another try then :)

Prove: (a*b)2 = e
(a * b) * (a * b) = LS
a * (b * a) * b = LS
a * (a * b) * b = LS
(a * a) * (b * b) = LS
(e) * (e) = LS
e = LS = RS
Therefore, H is closed under *

Is this correct?

11. Mar 18, 2012

### jgens

Yep. I am assuming that LS means left-hand side and RS means right-hand side.

12. Mar 18, 2012

### sunjin09

There's a theorem that says subset H is a subgroup of G iff for any α, β in H, α*β^{-1} is in H. I think this would simplify your proof.