I am a math hobbyist studying Joseph Gallian's "Contemorary Abstract Algebra" (Fifth Edition).
Part of Theorem 6.2 on page 124 states that if is an isomorphism then the order of an element a is equal to the order of ; that is (that is isomorphisms preserve orders).
Gallian's proof is essentially as follows:
a has order n
Gallian seems to conclude the proof at this point, but to me he has only shown (using his Th 4.1 Corollary 2) that the order of divides n
How do you show that the order of actually equals n?
put another way, φ(a^n) = (φ(a))^n. now if (φ(a))^m = φ(a^m) = e', that means both e = a^n, and a^m ≠ e are in the kernel of φ, meaning φ is not injective, and thus could not be an isomorphism.
in fact, φ does not need to be an isomorphism for this to be true, only a monomorphism (injective homomorphism), that is, it is injectivity (1-1) that preserves orders. here's an example:
consider the map φ:Z2 -->Z4 given by φ(k) = 2k. φ is injective. 0 has order 1 in Z2, and φ(0) = 0 has order 1 in Z4. 1 has order 2 in Z2, and φ(1) = 2 has order 2 in Z4. in this example φ preseves orders even though φ is not an isomorphism of Z2 with Z4.