We have then .

Thus, these two sets are the cosets.

First thus .2. Find the right cosets of the subgroup H = <(1,1)> in Z2 x Z4

These are the cosets.

Of course, since thus it is reflexsive. It is symmetric because if but then . Finally, it is transitive because if this means but then . This completes the proof that is an equivalence relation.3. Let H be a subgroup of a group G and define H congruent on G by letting x congruent to y IFF x^-1y is in H.

a) Show that H congruent is an equivalence relation on G

b) Show that the equivalence classes under H congruent are the left cosets of H in G.

c) Show that for a,b in G, aH = bH iff a^-1b is in H.

Now we wish to prove where and . Remember to show two sets are equal show that any element in is an element in and any element in is in . If and so for some , but then this means . The reserve direction is straightforward too.

Remember that so for we require that , and so .

Here is a hint for these problems. If is a finite group then .4. Let G = Q8 (quaternion group). Find [G : H] for H = <-I>, H = <K>, and H = <-L>.

5. a) In G = (Z48, +), find [G : H] for H = <32>.

(+ is actually addition mod 48, if it was unclear). There are two more parts to this question but if someone could tell me how to do this one, I can follow with the others.

This is Mine 12,2th Post!!!