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

Gallian's proof is essentially as follows:

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

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

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

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 $\displaystyle \phi (a)$dividesn

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