5
$\begingroup$

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.

$\endgroup$
6
  • 2
    $\begingroup$ Hint: Show that the order of $\phi(a)$ divides that of $a$. $\endgroup$
    – Soby
    Jan 3 at 22:27
  • 3
    $\begingroup$ $\ker \phi$ is trivial so $\text{im}\ \phi \cong G$ and obviously $|\phi(a)| \Big| |\text{im}\ \phi|$. $\endgroup$
    – 0XLR
    Jan 3 at 22:34
  • 1
    $\begingroup$ @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? $\endgroup$ Jan 3 at 22:37
  • 2
    $\begingroup$ 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 $\text{im}\ \phi \cong G$. $\endgroup$
    – 0XLR
    Jan 3 at 22:39
  • 2
    $\begingroup$ @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 $H$. $\endgroup$
    – D_S
    Jan 3 at 22:44
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\phi(G)$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.