2.4.11

Let G and H be cyclic groups generated by the elements x and y with orders m and n respectively (m,n > 0). Consider the map f from G to H defined by $$ f(x^i) = y^i $$. We claim that f is a homomorphism if and only if m is an integer multiple of n.

Suppose f is a homomorphism. Then $$ f(x^m) = y^m $$, but since x has order m, $$ x^m = 1 $$, thus $$ f(1) = y^m $$. By Proposition 2.4.3, homomorphisms map the identity to the identity, so $$ f(1) = 1 $$, from which $$ y^m = 1 $$. By definition, the order n of y is the smallest positive integer such that $$ y^n = 1 $$, so necessarily $$ m = kn $$ for some positive integer k.

Conversely suppose $$ m = kn $$ for some positive integer k. Let $$ x^a, x^b \in G $$, where $$ a,b \in \left\{0,1,...,m-1\right\} $$. Then, $$ x^a{}x^b = x^{a+b} = x^p$$, where p = a+b mod m, and $$ 0 \le p < m $$. (Such a p exists due to the Euclidean algorithm.) So, $$ f(x^a{}x^b) = f(x^p) = y^p$$, where $$a+b = sm + p $$ for some integer s. Now $$ m=kn $$, so $$a+b = skn + p $$, thus a+b = p mod n.

But also, $$ f(x^a)f(x^b) = y^a{}y^b = y^{a+b} = y^q $$, where q = a+b mod n and $$ 0 \le q < n $$. But since a+b = p mod n, we have p = q mod n , or $$ p = ln + q $$ for some integer l, so therefore $$ y^p = y^q$$, and hence $$ f(x^a{}x^b) = f(x^a)f(x^b) $$, showing that f is a homomorphism. Hence f is a homomorphism if and only if $$ m = kn $$ for some positive integer k.