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Math Help - Subgroup of order 4

  1. #1
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    Lightbulb 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?
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  2. #2
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    Re: Subgroup of order 4

    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.

    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.
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  3. #3
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    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
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  4. #4
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    Re: Subgroup of order 4

    Your reasoning for the first part is ok.

    Quote Originally Posted by MAX09 View Post
    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.
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