Using counting formula to get |G| = |kernel φ||image φ|

Question

The counting formula I am saying :

Let S be a finite set on which a group G operates, and let Gs and Os be the stabilizer and orbit of an element s of S. Then |G|=|Gs||Os| or (order of G)=(order of stabilizer)(order of orbit)

then I have the following question:

Suppose that φ : G → G′ is a homomorphism between finite groups. Explain how the familiar formula |G| = |kernel φ||image φ| follows from the counting formula.

I was told that if G acts on the set of left coset of Kernel φ, i.e. G/K,

then by counting formula, use |kernel φ|=|stabilizer| and |image φ|=|orbit|

we can get |G| = |kernel φ||image φ|

but I can not rationalize it, I think it is due to my poor understanding of coset. Can someone explain it a little? Thanks in advance.

Answer 1

(Left) cosets are subsets of $G$ of the form $gK$ (here with $K=\ker \phi$). For $g,h\in G$ we have either $gK=hK$ (even if possibly $g\ne h$ of $gK\cap hK=\emptyset$, and the union of all cosets is of course all of $G$ (as $g\in gK$).

How does $G$ act on $G/K$? For $g\in G and $A\in G/K$ (say, $A=hK$ with $h\in G$) the set $gA=ghK$ is also a left coset - and that's it.

Let us pick a specific coset in $G/K$ and determine ists stabilizer. For simplicity, pick $K$ itself. For which $g\in K$ is $gK=K$? Since $e\in K$, this specifically implies $g\in K$, i.e. $g\in\ker \phi$ by our choice of $K$. On the other hand $g\in K$ also gives us $gK=K$. So the stabilizer of $K$ is just $\ker \phi$ itself.

How can we relate $\operatorname{im}\phi$ with $G/K$? If $g'\in G'$ is in the image of $\phi$, then $\phi^{-1}(\{g\})$ is a left coset (if $\phi(g)=g'$ and $k\in G$ then $\phi(gk)=g'$ iff $k\in\ker\phi$) and of course for different $g'$ we obtain different cosets. (On the other hand if $g'$ is not in the image, then $\phi^{-1}(\{g\})=\emptyset$) Thus $\phi$ induces a bijective map between the sets $G/K$ and $\operatorname{im}\phi$ (in fact this is a group isomorphism, but we don't need that here; one could do this part with an isomorphism theorem anyway)

Answer 2

You don't need to explicitly use cosets to reach this equality.

Let $G$ act on set $G'$ using the rule $gg' = \varphi(g)g'$ for every $g \in G$ and every $g' \in G'$. It is a trivial exercise to check that this is indeed a group action.

Now you just look at the orbit of the identity element $1_{G'} \in G'$. Its orbit is precisely the image of $\varphi$, and its stabilizer is the kernel of $\varphi$, and you are done.