Thread: 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.

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.

Thanks for your help Deveno. I was able to figure it out. I used S3 as the counter-example.