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Thread: Arbitrary vs. Specific Elements (Coset)

  1. #1
    Junior Member
    Oct 2010

    Arbitrary vs. Specific Elements (Coset)

    Here is the problem I am given:

    Let H be a subgroup of G and let a, b, x, y be in G. Prove or disprove:
    "If a*H = b*H then a^2*H = b^2*H"

    so I let the set a^2*H be {a^2 * h: h in H}
    since a is in group G, a * a is in G
    let a * a = c
    similarly, let b * b = d in G
    so we have the sets a^2*H = c*H and b^2*H = d*H

    this is where my question comes into play. are a and b specific elements, such a*H = b*H does not say anything about c*H and d*H? Or can we now say c*H = d*H because of a and b.

    normally i don't have any trouble with this issue, but i'm getting tripped up here for some reason.
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  2. #2
    MHF Contributor

    Mar 2011
    if aH = bH, this means that for h in H, ah = bh' (for some different element h' of H).

    now if we look at a^2H, we have the element a^2h = a(ah) = a(bh').

    from ah = bh', we can conclude that a = bh'h^-1, so a^2h = (bh'h^-1)(bh').

    the problem is, b may not commute with elements of H, so we cannot say if bH = Hb.

    if fact, if bH ≠ Hb, we can't take the second b in bh'h^-1bh' "across" the h'h^-1 part to get it with the first b to make it be b^2(stuff in H).

    now, formalize this, by finding a small group G, with a subgroup H and an element g with gH ≠ Hg.

    you'll want a non-abelian group. then find some other element g' of G, with g'H = gH, and (g')^2H ≠ g^2H.
    Last edited by Deveno; Mar 29th 2011 at 04:26 PM.
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  3. #3
    Junior Member
    Oct 2010
    Thanks for your help Deveno. I was able to figure it out. I used S3 as the counter-example.
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