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**Problem 1: Let G be a group generated by elements a, b with a^4 = e, b^2 = e, baba = e. Show that G has at most 8 elements. Is there such a group with precisely 8 elements?**

Problem 2: Let H, K be normal subgroups of a group G whose intersection contains only e_G, and whose union generates the whole group G. Prove that G is isomorphic to H x K. (Hint: Prove that hk = kh for all h in H and k in K, and consider the map H x K --> G with (h,k) |--> hk.)

Problem 2: Let H, K be normal subgroups of a group G whose intersection contains only e_G, and whose union generates the whole group G. Prove that G is isomorphic to H x K. (Hint: Prove that hk = kh for all h in H and k in K, and consider the map H x K --> G with (h,k) |--> hk.)

with 1, it's easy to show that G has at most 8 elements. I'd simply write them all out, and come across some congruent elements. So we'd have the following group

e

a

b

ab=ba

aa

bb=e -----> shows b cannot generate anymore distinct elements

aab

aaa

aaab

aaaa=e

However, is there a group with exactly 8 elements? Thinking about it...we'd have to have exactly one representation of each value. If each is to have it's own inverse, then thered have to be an even amount of elements, excluding the neutral. This means that either an odd amount of elements are their own inverses, or that theyre all all theyre own inverses. I believe that it doesn't work out , but how can i prove it? Any help as for where to go?

As for 2...im not sure where to begin either, even though the hint is there. I will keep you posted with my progress however. thank you in advance