Show that two infinite cyclic groups are always isomorphic.
Let G and H be two infinite cyclic groups generated by elements a and b respectively.
Since G is generated by the element a, all elements in G can be written as a^k, where k is an integer. Similarly, since H is generated by the element b, all elements in H can be written as b^j, where j is an integer.
Define a function f: G -> H such that f(a^k) = b^k for all integers k.
We claim that f is an isomorphism between G and H.
1. f is a homomorphism:
f((a^m)(a^n)) = f(a^(m+n)) = b^(m+n) = b^m * b^n = f(a^m) * f(a^n)
2. f is onto:
Let h be an arbitrary element in H. Since H is generated by b, h = b^j for some integer j. Then, f(a^j) = b^j = h, so f is onto.
3. f is one-to-one:
Suppose f(a^m) = f(a^n) for some integers m and n. Then, b^m = b^n, which implies that m = n. Therefore, f is one-to-one.
Since f is a homomorphism that is onto and one-to-one, f is an isomorphism. This shows that G and H are isomorphic to each other, and hence any two infinite cyclic groups are always isomorphic.