# $|G| =(\text{order of image of} f) |H|$

I am learning groups in abstract algebra. My current topic is order of groups and Lagrange's Theorem. Here is the version of the Lagrange's Theorem that I learnt:

Let $S$ be a finite group and $H$ a subgroup of $S$. Then, $|H|$ divides $|S|$.

Here is the question that I came over:

Let $f: G \rightarrow G'$ be a homomorphism with kernel $H$. Assume that $G$ is finite. Show that $$|G| =(\text{order of image of} f) |H|.$$

It is simple to verify that $H$ is a subgroup of $G$. Thus, $|H|$ divides $|G|$. But how would I relate this division by the order of the image set?

Hint: In the proof of Lagrange's theorem, you show that $$|G| = (G:H)\ |H|$$ where $(G:H)$ is the number of left cosets of $H$ in $G$.

Can you find a bijection (not a homomorphism) between the left cosets of $H$ in $G$ and the image of $f$?

Extra hint:

Can you show that the function that takes $gH \mapsto f(g)$ gives a well-defined bijection?

Two elements of $G$ have the same image under $f$ iff they are in the same coset of $H$.

In other words, the inverse image of an element of the image of $f$ (also called a fiber of $f$) is a coset of $H$. The number of fibers is clearly the size of the image of $f$.

Since every element of $G$ is in exactly one fiber of $f$, we get the result.