Subgroup of order 4

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March 20th 2013, 10:57 PM
MAX09

Subgroup of order 4

I just wanna know if the following proof is okay for my assignment problem.

Question: Prove that an abelian group with 2 elements of order 2, must have a subgroup of order 4.

Proof: Let a,b belong to G such that a² = b² = e.

Case:1 G is finite

Now, I propose that there should atleast one more element,x in G.
This is true according to a corollary of Lagarage's theorem which states that,
"In a finite group, the order of the an element of the group divides the order of the group"

So, o(G) = 2n (n>1)

Case 1a Let's first prove there is one more element in G.

We know a² = b² = e. Consider the order of ab.
(ab)² = a * b * a *b = a* a * b *b = a² * b² = e.
So, the other element in the group, say x is the product of a and b.
So, G = {e,a,b,x}, When x is denoted as c, this becomes the Klein's 4 group, eh?
Now, every group is a subgroup of itself, the result is proved.

Case 1b Let's suppose there are more than 4 elements in G. Let o(G) = 2n (n>2)
Still, this set H = {e, a, b, ab} satisfies all the necessary conditions for becoming a subgroup of G which proves that G still has a subgroup of order 4.

In addition, i believe this also proposes that the order of G is 4n (n>=1), as any other order would be a contradicting the Lagrange's theorem.

I'm a little confused about the proof for Case 2, where G is of infinite order. We cannot use the corollary due to Lagrange's theorem. Can someone help me with that?
March 20th 2013, 11:59 PM
Gusbob

Re: Subgroup of order 4

Quote:

We know a² = b² = e. Consider the order of ab.
(ab)² = a * b * a *b = a* a * b *b = a² * b² = e.
So, the other element in the group, say x is the product of a and b.
So, G = {e,a,b,x}, When x is denoted as c, this becomes the Klein's 4 group, eh?
Now, every group is a subgroup of itself, the result is proved.

I know it seems obvious, but you should check that $ab\not= e$ . What you have shown is that $ab$ has order 1 or 2.

Quote:

Case 1a Let's first prove there is one more element in G.

Is there anything wrong with your particular argument when $G$ is any other (possibly infinite) abelian group? Just use the fact that $a\in G,b\in G \implies ab\in G$ to get rid of your restriction on the size of $G$ .
March 21st 2013, 06:22 AM
MAX09

Re: Subgroup of order 4

Thanks, i never took ab=e into account, seriously. This is cos i thought ab = e ===> a and b are inverses of each other, but since a² = e, we see that a is the inverse of itself, rather an idempodent element. the same goes for b, therefore, i guess ab =e would not be true. is my point strong enough?

About, the part about, the infinite abelian group, i couldn't get to the result which uses the same logic that you suggested, namely, "a and b belonging to G ===> ab also belongs to G"

Can you help me with that part alone?

thanks, Gusbob
March 21st 2013, 05:37 PM
Gusbob

Re: Subgroup of order 4

Your reasoning for the first part is ok.

Quote:

Originally Posted by MAX09

About, the part about, the infinite abelian group, i couldn't get to the result which uses the same logic that you suggested, namely, "a and b belonging to G ===> ab also belongs to G"

I'm not sure what you're asking here, but I'll answer both of my interpretations of it.

If you're asking why $a,b\in G \implies ab\in G$ , that is just the definition of a group: A group is a set $G$ with a law of composition. Formally, a law of composition on a set $S$ is a function $S\times S \to S$ . So you have to get an element in your set again.

If you're asking how your proof generalises, here is a short proof that works for all cases (not well written, however, I'm a bit in a hurry, sorry):

Suppose $a$ and $b$ have order 2 in $G$ . Then $ab$ is also an element of $G$ , order 2 (because of the reasoning you gave). The claim now is that $H={e,a,b,ab}$ is a subgroup of order 4. Since it has 4 elements, we really only need to check that it is a subgroup:

- Contains an identity: yes
- closed under composition: $a\cdot b= ab\in G, a\cdot ab= b, b\cdot ab = a$ . We omit the obvious relations with $e$ , and the other non-trivial relations are covered by the commutativity of the group $G$
- Inverses: each element is its own inverse, so yes.