1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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


    User Avatar
    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


    User Avatar
    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


    User Avatar
    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook