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

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.

• Hint: Show that the order of $\phi(a)$ divides that of $a$.
– Soby
Jan 3 at 22:27
• $\ker \phi$ is trivial so $\text{im}\ \phi \cong G$ and obviously $|\phi(a)| \Big| |\text{im}\ \phi|$.
– 0XLR
Jan 3 at 22:34
• @0XLR Do you mind writing a definition of $im\phi$ please? I am confused on I thought that reads image of phi hence would mean $H$, but we are trying to show order of $G$. Is this the same thing since phi is one-to-one? Jan 3 at 22:37
• As I said, the image of $\phi$ is isomorphic to $G$; so it has the same order as $G$. And yes, that is because of the one-to-one property: the kernel is trivial and the first isomorphism theorem gives $\text{im}\ \phi \cong G$.
– 0XLR
Jan 3 at 22:39
• @NormanContreras the definition of $\operatorname{im}\phi$ is $\{ \phi(g) : g \in G\}$. It is also denoted by $\phi(G)$, and is a subgroup of $H$.
– D_S
Jan 3 at 22:44

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.
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.
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 $$\phi(G)$$?