3
$\begingroup$

I was reading Group Homomorphisms from Contemporary Abstract Algebra - Gallian. In pg-216, 8th ed (pg-200, 4th ed), the author writes -

"We know that the number of homomorphic images of a cyclic group G of order n is the number of divisors of n, since there is exactly one subgroup of G (and therefore one factor group of G) for each divisor of n. (Be careful: The number of homomorphisms of a cyclic group of order n need not be the same as the number of divisors of n, since different homomorphisms can have the same image.)"

How are homomorphic images and homomorphisms different? They sound very similar... Thanks..

$\endgroup$
  • $\begingroup$ There are five different homomorphisms from $C_5$ to $C_5$, namely $x\mapsto x$, $x\mapsto x^2$, $x\mapsto x^3$, $x\mapsto x^4$, and $x\mapsto x^5$, but four of them have the same image. (There's no way they could have five different images, because each image is a subgroup of $C_5$, and there are only two different subgroups.) $\endgroup$ – bof Dec 3 '14 at 11:27
4
$\begingroup$

Let $\phi:G\rightarrow H$ be a mapping from groups $G$ to $H$ .If $\phi$ satisfies $\phi(ab)=\phi(a)\phi(b)$ for any $a,b\in G$ then $\phi $ is said to be a group homomorphism from $G$ to $H$ and $\phi(G)\subseteq H$ is said to be a homomorphic image of $\phi$.Moreover if $\phi$ is surjective then $\phi(G)=H$ and $H$ is a homomorphic image of $G$ under $\phi$

$\endgroup$

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.