Let $G,H$ be groups and $\phi\colon G\to H$ be a one-to-one group homomorphism. Let $a\in G$. Show that $|\phi(a)| \mid |G|$. Not sure how to attack this proof.
Do we use the First Isomorphism Theorem and Lagrange's Theorem? I am unsure if the canonical homomorphism is helpful for this proof.
Best Answer
By Lagrange, the order of $a$ divides the order of $G$. By the homomorphism property, the order of $\phi(a)$ divides the order of $a$. Now by transitivity of divisibility, the result follows.
You don't need injectivity.
However, injectivity could be considered to simplify the proof. Because $\phi(a)$ will be an element of the image. Then apply the first isomorphism theorem and Lagrange.