Show that $|\phi(a)| \mid|G|$

Question

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.

Answer 1

Let $\phi(G) = \{ \phi(g) : g \in G\}$ be the image of $G$ under $\phi$. It is a subgroup of $H$ (check this), and in particular, $\phi(G)$ is a group. Because $\phi(G)$ is a group containing the element $\phi(g)$, Lagrange's theorem tells us that the order of $\phi(g)$ divides the order of $\phi(G)$.

Now because $\phi$ is assumed to be one-to-one, what does that tell you about the relationship between the order of $G$ and the order of $\phi(G)$?

Answer 2

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.