Let a, b, c, d be integers, and let H be the subgroup of Z × Z generated by (a, b) and (c, d). Thus H is the set of all elements of the form (ma, mb) + (nc, nd) where m, n ∈ Z.
(a) Suppose b = 0, c = 0. Prove that (Z × Z)/H is isomorphic to Z/aZ × Z/dZ.
(b) Suppose (a, b) = (10, 12) and (c, d) = (4, 4). Find a product of cyclic groups isomorphic to (Z × Z)/H .
(c) Determine, in terms of a, b, c, d, when (Z × Z)/H is ﬁnite. When it is ﬁnite, ﬁnd its order.
I can see intuitively that the statements for a and b are right but I can't come up with any method of proof
could anyone help me?? plelase ?
Sep 24th 2009, 12:58 PM
For (a), construct a surjective (onto) homomorphism from to whose kernel is , then apply the First Isomorphism Theorem for groups.
For (b), recall that 'modding' out by H just means we impose the relation (10m+4n,12m+4n)=(0,0) in . In which subgroup of does 10m+4n=0 for all m,n? 12m+4n=0? The product of these two groups is what you are looking for.
For (c), look at gcd(a,b) and gcd(c,d), and recall that the gcd(x,y) can always be written as a linear combination of x and y.