Verified answer

1). The relation ᴴ≡ is an equivalence relation if it satisfies the following:

- x ᴴ≡ x

- if x ᴴ≡ y then y ᴴ≡ x

- if x ᴴ≡ y and y ᴴ≡ z, then x ᴴ≡ z.

- Because the identity element is in H because H is a subgroup, (x^-1)x is in H, so x ᴴ≡ x

- Say x ᴴ≡ y. Then we know (x^-1)y is in H. Now because H is a subgroup, its inverse is also in the group. Note that

[(x^-1)y ]^-1 = y^-1x. So we have (y^-1)x is in H. So y ᴴ≡ x.

- Say that x ᴴ≡ y and y ᴴ≡ z. Then (x^-1)y is in H and (y^-1)z is in H. So their product is also in H:

(x^-1)y*(y^-1)z = (x^-1)z is in H. So x ᴴ≡ z.

So ᴴ≡ Is an equivalence relation.

2). For a given element g in G, the equivalence class of g, [g] consists of all the x in G that satisfy

(g^-1)x is in H. This happens precisely when x is of the form gh, where h is in H. So the conjugacy class of g, is the set gH, which is the left coset of H determined by g. So the equivalence classes are just the left cosets gH.

3). We have to show two implications (iff is if and only if!). Sorry for the long hold, my internet failed XD

Anyways :

assume aH = bH. Multiplying both sides with b^-1, we obtain

(b^-1)aH = H, which implies that (b^-1)a is in H. Thus its inverse is also in H, because H is a subgroup. The inverse is:

[(b^-1)a]^-1=(a^-1)b. So (a^-1)b is in H and we've shown the first implication.

Now assume (a^-1)b is in H. Then we have (a^-1)bH = H, so multiplying on the left with a gives us

bH = aH. So we've shown the reverse implication as well.

3) suppose aH = bH. then for any h in H, ah = bh' for some other h' in H.

thus h = a^-1bh', and so:

a^-1b = hh'^-1, which is in H

on the other hand, if a^-1b is in H, then a^-1b = h, for some h in H.

let h' be any element of H.

then bh' = a(hh'), so bH is contained in aH.

also, ah' = b(h^-1h'), so aH is contained in bH

(or we can use that fact that two cosets either coincide,

or are disjoint to show that bH is contained in aH to show

that aH = bH).