Property on homomorphism between finite groups

Question

I was reading some lecture notes on homomorphism between finite groups and intuitively it appeared to me that for a $\phi$ an homomorphism $G \rightarrow H$, we should have:

$$|\,\textrm{Im}(\phi)\,| = \frac{|\,G\,|}{|\,\textrm{ker}(\phi)\,|}$$

where $| . |$ stands for order. Is this right? If yes, is there a short demonstration for?

Answer 1

Yes. Recall that that given a group homomorphism $\phi \colon G \to G'$, $\ker \phi$ is a normal subgroup of $G$. The first isomorphism theorem states that given such a homomorphism, there is an isomorphism $G / \ker \phi \cong \text{Im} \phi$. This gives your result. A proof can be found in any elementary book on Abstract Algebra, or by simply Googling it.

Answer 2

ker($\phi$) is a normal subgroup of G, so G/ker($\phi$) is well defined and |G/ker($\phi$)|=|G|/|ker($\phi$)|. Thanks to "the first isomorphism theorem", there exists an isomorphism between Im($\phi$) and G/Ker($\phi$) moreover |Im($\phi$)|=|G|/|Ker($\phi$)|