2.6.2

Prove directly that distinct cosets do not overlap.

This is a proof of the contrapositive, not a direct proof. Let G be a group, H be a subgroup of G, and let aH and bH be two left cosets of H with nonempty intersection. If $$a=b$$, then the cosets are evidently equal and we are done. Otherwise, there exists an element g in both aH and bH, such that $$g = ah$$ for some h in H, and also $$g = bh'$$ for some h' in H, so that $$ah = bh'$$, implying $$a = bh'h^{-1}$$ and $$b = ahh'^{-1}.$$

Now let $$ah_1$$ be an element of aH, where $$h_1$$ is in H. Then, $$ah_1 = bh'h^{-1}h_1 $$, and since $$h',h^{-1},h_1 $$ are all in the subgroup H, so is their product, so that $$bh'h^{-1}h_1 $$ is in bH, showing aH is contained in bH. Similarly, given $$bh_2 $$ in bH, we have $$bh_2 = ahh'^{-1}h_2$$, showing $$bh_2$$ is in aH, so that aH = bH.

The proof in the case where the cosets are right cosets (Ha and Hb) of H is essentially the same: $$ha=h'b $$ for some $$h, h' $$ in H, letting us relate a and b with $$a = h^{-1}h'b$$ and $$ b = h'^{-1}ha $$, from which $$h_1a=h_1h^{-1}h'b$$ and $$h_2b=h_2h'^{-1}ha $$, showing Ha and Hb are contained in each other and hence equal. Thus if two cosets overlap, they are equal.