# Thread: Isomporhisms preserve orders - help with proof

1. ## Isomporhisms preserve orders - help with proof

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 $\phi :G \rightarrow G'$ is an isomorphism then the order of an element a $\in G$ is equal to the order of $\phi (a)$ ; that is $\ \mid a \mid \ = \ \mid \phi (a) \mid$ (that is isomorphisms preserve orders).

Gallian's proof is essentially as follows:

a has order n $\ \ \Longrightarrow a^n = e$

$\ \Longrightarrow \ \phi (a^n) \ = \phi (e) \ = \ e'$

$\ \Longrightarrow \ [\phi (a)]^n \ \ = \ e'$

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 $\phi (a)$ divides n

How do you show that the order of $\phi (a)$ actually equals n?

2. Originally Posted by Bernhard
How do you show that the order of $\phi (a)$ actually equals n?

If $( \phi (a))^m=e'$ for $m then ... (absurd).

3. Fernando, why exactly is it absurd ... obviously I am missing something

Can you be more explicit?

[I am aware that the equality is not necessarily true for homomorphisms so it must be something to do with 1-1 and onto]

Bernhard

4. $\phi^{-1}$ is an isomorphism so if $(\phi(a))^{m}=e'$ taking $\phi^{-1}$:

$\phi^{-1}[\;\phi(a)^m\;]=\phi^{-1}[\;e'\;]\Rightarrow a^m=e$

The order of $a$ wouldn't be $n$ .

5. Thanks Fernando

6. Originally Posted by Bernhard
Thanks Fernando

You are welcome.

7. 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.