Let H = {x in group G | x^2 = e}. Prove or disprove that....

[tegra97]

Let G be a group with identity e. Let H = {x belongs to G | x^2=e}
Prove or disprove H is a subgroup of G.

I'm having some trouble with this problem, I'm looking in my book and I see subgroup tests such as the one-step subgroup test, two-step subgroup test, and finite subgroup test. Should I use one of these test to prove it? Also in the text it says that a group is not a subset if any one of the 3 happen.

1. show that the identity is not in the set.
2. Exhibit an element of the set whose inverse is not in the same set.
3. Exhibit two elements of the set whose product is not in the set.

Would this be the easiest way to prove it?
Thanks a bunch.

 

[pka]

In order to disprove a subset of a group is not a subgroup, one must have a particular group to work with. This problem has no particular group.

 

[tegra97]

So you're saying that H is not a subgroup of G, because the elements commute.

 

[pka]

No, I said quite the opposite. Here is a theorem.
A set H of a group G is a subgroup of G if and only if when a & b are in H then ab<SUP>-1</SUP> is also in H.

Can you prove that H is a subgroup in this case?

 

[mcwang719]

so can we say a^2=e and b^2=e so (ab^-1)=e. So (ab^-1)^2= ab^-1ab^-1= a^2(b^-1)^2= (a^2)(b^2)^-1= ee^-1= ab^-1.

 

[daon]

so can we say a^2=e and b^2=e so (ab^-1)=e.

How did you reach this conclusion?? It might be a good idea to justify every step or at least make obvious transitions to your conclusion. What you wrote imples a=b.

(ab^-1)^2= ab^-1ab^-1= a^2(b^-1)^2= (a^2)(b^2)^-1= ee^-1= ab^-1.

So, (ab^-1)² = ab^-1? You should have stopped at "=e." Also, do you know the elements commute? If it hasn't been proven, your professor may expect you to do it in a little lemma.

 

[mcwang719]

Sorry let me rephrase that. What I got from the book is we assume a and b belong to H. This means a^2=e and b^2=e and we must show that (ab^-1)=e.

So (ab^-1)^2= ab^-1ab^-1= a^2(b^-1)^2= (a^2)(b^2)^-1= ee^-1=e.

This is to prove H is a subgroup of G, is that correct?
What if G is abelian and H is a subgroup?

 

[daon]

Sorry let me rephrase that. What I got from the book is we assume a and b belong to H. This means a^2=e and b^2=e and we must show that (ab^-1)=e.

Isn't the goal to show \(\displaystyle ab^{-1} \in H\)? So you don't WTS \(\displaystyle ab^{-1}=e\) but rather \(\displaystyle (ab^{-1})^2=e\)

This is to prove H is a subgroup of G, is that correct?
What if G is abelian and H is a subgroup?

Yes, that is correct. If G is abelian, then H is abelian if it is a subgroup of G. But we can't use that. Here, unless we are given G or given that G is abelian, it is impossible to tell if G is abelian. We can only tell if H is abelian. If in fact it is that must also be proved if you want to use it using the definition of H.

You used the fact that H is abelian in your proof, so in order for your proof to be valid you need to show it.

 

[mcwang719]

There are two parts to this question. Prove or Disprove:
a) H is a subgroup of G
b) If G is abelian, H is a subgroup of G

I'm a little confused, so I just proved a) and how is proving b) any different from a)? Thanks.

 

[daon]

You haven't proven a. You proved b:
G abelian => (ab^-1)²=a²(b^-1)²=a²b²=ee=e.

IF you successfully proved (a), part (b) would be free.

Do you NEED commutativity for H to be a subgroup of G? If so, is H an abelian group? That is what you need to figure out.

 

[mcwang719]

Yes you need commutativity of H to be a group. So if G is non-abelian than, than H is not a subgroup?

 

[daon]

Exactly (kind-of). When trying to show closure of H we get the following: If a,b are in H then

\(\displaystyle (ab)^2 = (ab)(ab)\)

If G is abelian then \(\displaystyle (ab)(ab) = a^2b^2 = e\). Thus \(\displaystyle ab \in H\).

We needed that assertion to continue.

However, that doesn't prove that G has to be abelian, only that you needed it for this proof. You might want to try thinking of a counter-example if you need to disprove it. I'm not sure of one off-hand. Best bet is to think of some matricies since you know they don't generally commute.

 

[daon]

I spent a good hour on this, lol. I found a counter example for you. (I just played a minute more and found a much more simple GL(2,R) example with 2x2 matricies)

\(\displaystyle \L GL_3(\mathbb{R})\) is the group of 3-by-3 real-entry matricies that have a non-zero determinant. That this is a group should be verified unless you've already done it in class or something.

\(\displaystyle \L

H = \{ A \in GL_3(\mathbb{R}) \,\, \| \,\, A^2 = I \}\)


Goal: show that \(\displaystyle \L H\) is not a subgroup of \(\displaystyle \L \,\, GL_3(\mathbb{R})........\)

\(\displaystyle \L A = \[ \begin{array} -1 && 0 && 0 \\ 0 && -1 && 0 \\ 0 && 0 && 1 \end{array} \] \\\)

\(\displaystyle \L B = \[ \begin{array} -1 && 0 && -1 \\ 0 && -1 && 0 \\ 0 && 0 && 1 \end{array} \] \\\)



Now, I will leave it to you to verify that:

\(\displaystyle \L
A^2 =I, \,\, \,\, B^2=I, \,\, \,\, \text{but} \,\, \,\, (AB)^2 \neq I\)

 

[mcwang719]

Thank you very much Daon I really appreciate it! You have been extremely helpful!