group theory - indirect product group?

Question

I am stuck on this question, and i know what i need to do, i just don't really know how to do this.

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

I know that to do this i need to show that G is an indirect product group of A and B. This means i need to show:

1. A and B are normal subgroups of G
2. A intersection B = the identity of G
3. G = AB (i.e. for all g in G there exists an a in A and a b in B such that g = ab)

I have done the easy part of proving point 2, but i am not sure how to go about proving points 1 and 3. Any help would be greatly appreciated.

Answer 1

You don't need to use your criteria. To show two groups are isomorphic, you just have to show that there is an isomorphism. It is true that the product is an internal direct product (Your #3), but unless your question asked for that specifically, you don't need that. Let G be the cyclic group generated by g. Let A be generated by a and B be generated by b. Here is a map from G to A x B:

g maps to (a,b)

It is trivial to check that this is a homomorphism, it is injective, and it is onto (recall that p and q are *distinct* primes, i.e. p and q are relatively prime).

If you must show that the product is an internal direct product and then use that to show the direct product, then do this:

For number 1, use the fact that G is an Abelian group. Every subgroup of an Abelian group is normal (what does conjugation do in an Abelian group?).

For number 3, G is isomorphic to Z/pqZ, where Z denotes the integers and the group operation is *addition* (note that, as written, G is multiplicative). This just simplifies our notation, and anything we say about Z/pqZ can be translated back to G via the isomorphism (i.e. the inverse image of the element 1 generates G, the inverse image of q in G generates A, etc.) I'm going to just use the isomorphism and say that G=Z/pqZ , A=

Answer 2

some suggestions of the thank you to attack it: a million. See what the Sylow Theorems say. 2. locate cyclic subgroups of orders 2, 3, 4, and 5. Take their products and notice what the orders of the ensuing communities are. :) 3. answer a similar question for A4, embedded in A5 interior the glaring way. For those communities, take their products with the communities generated by skill of by skill of (40 5) and (12345).

Answer 3

1. Suppose a^m is a member of A (or B), then for every g in G there is n such that g=a^n, (g^-1) * a^m * g = (a^-n)a^m(a^n) = a^(-n+m+n)=a^m.

3. p and q are prime numbers, so there are two integers namely m and n such that mp + nq = 1.
a^(mp) is a member of A, and a^(nq) is a member of B. (For every integer because A and B are sub-groups).
Now, a^k is a member af g that can be written as a^(mpk) * a^(nqk).

QED.