Group theory: Direct products?

Question

I dunno where to start on this question, any help would be greatly apprieciated.

Let G be a cyclic group of order pq where p, q are distinct primes. Then G = for some a with order pq. Show that A = which is isomorphic to Z_p (the set of integers modulo p) and B = which is isomorphic to Z_q (the set of integers modulo q) and G is isomorphic to the direct product A x B.

I dont really understand what or means, so if someone can show me how to prove the part about A, i think i would be able to complete the rest of the question. Thanks.

Answer 1

To say G= means that there is an element a in G so that every element of G is a power of a. To say a has order pq means that a^{pq}=e, and that this is the first power which is the identity. The problem asks you to consider the set of powers of a^q. This will be a subset of the powers of a, and thus is a subset of G. It is in fact a subgroup. To show that A is isomorphic to Z_p takes some work to do from scratch, and I'm not sure what you've seen before. I will use the fact that every group with a prime number of elements is isomorphic to Z_p. In fact A has p elements, as you can see because it must have {1,a^q,a^{2q},...,a^{(p-1}q}}, and the next power would be a^{pq}=e again, and the powers repeat. Counting these elements, we are done.

Answer 2

That notation looks like it has something to do with generators.

Answer 3

WTS: ?((g_1, g_2) + (g_3, g_4)) = ?(g_1, g_2) + ?(g_3, g_4) ?((g_1, g_2) + (g_3, g_4)) = ?(g_1+ g_3, g_2 + g_4) = (f(g_1 + g_3), g(g_2 + g_4)) = (f(g_1) + f(g_3), g(g_2) + g(g_4)) via fact g and h are isomorphisms = (f(g_1), g(g_2)) + (f(g_3), g(g_4)) = ?(g_1, g_2) + ?(g_3, g_4) as a result, ? is an homomorphism. you could coach it rather is a bijection.