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Proof: H is a subgroup of G

  1. Mar 17, 2012 #1
    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. jcsd
  3. Mar 17, 2012 #2
    Careful here: you are confusing H={g ε G: g^2=e} with a different set, {g^2: g ε G}.
     
  4. Mar 17, 2012 #3
    g2*g2 = e
    g2 + 2 = e
    g4 = e
    (g2)2 = e
    Therefore, H is closed under *

    does this fix the issue?
     
  5. Mar 17, 2012 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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 [itex]a^2= 1[/itex] and [itex]b^2= 1[/itex]. So what can you say about [itex](ab)^2[/itex]?

    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.
     
  6. Mar 17, 2012 #5
    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
    ?
     
  7. Mar 18, 2012 #6
    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.
     
  8. Mar 18, 2012 #7

    jgens

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    Gold Member

    Let G = D6 and notice that s2 = e and that (sr)2=(sr)(r-1s) = e. But s(sr) = r and r2 ≠ e.
     
  9. Mar 18, 2012 #8
    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 *
    ?
     
  10. Mar 18, 2012 #9

    jgens

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    Gold Member

    You are not trying to prove that [itex]a,b \in H[/itex] implies [itex]ab = e[/itex]. You are trying to prove that if [itex]a,b \in H[/itex], then [itex](ab)^2 = e[/itex]. 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 [itex]G[/itex] 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 [itex]G[/itex] is non-abelian in order to show that [itex]G[/itex] abelian was necessary.
     
  11. Mar 18, 2012 #10
    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?
     
  12. Mar 18, 2012 #11

    jgens

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    Gold Member

    Yep. I am assuming that LS means left-hand side and RS means right-hand side.
     
  13. Mar 18, 2012 #12
    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.
     
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