Let $G$ a finite group and $N$ a subgroup. Show that $|G|=|N|\cdot|G/N|$

Question

Please, can you check and correct my proof? Note: the multiplication is used to symbolize the group operation, and $G/N$ is a partition of $G$ defined by the equivalence relation

$$a\sim b\iff a\in bN$$

where $bN=\{bx:x\in N\}$

My attempt at the moment: if $|G|=|N|\cdot |G/N|$ then must exist a bijection

$$f:G\to N\times G/N$$

Then I tried with a map of the kind

$$g\mapsto \langle n,[g]\rangle$$

where $n$ can be used as an enumeration, i.e. we can make a bijection $g:N\to\{1,...,m\}$ and change $n$ by $g(n)$, supposing that any equivalent class $[g]$ have exactly $m$ members ($|N|=m$).

To show that any equivalent class $[g]$ have cardinality $m$ it is enough to show that cannot exist $ab=ac$ for any $a,b,c\in G$:

$$ab=ac\to abb^{-1}=acb^{-1}\to ae=acb^{-1}\to\\ a^{-1}ae=a^{-1}acb^{-1}\to e=cb^{-1}$$

what cannot hold if $b\neq c$ due to the uniqueness of the inverse (the element $e$ represent the identity of the group operation).

Then from here is obvious that $f$ is injective, and cause the equivalent classes are a partition of $G$ then $f$ must be surjective, so $f$ is injective.

Questions:

1. It is my proof correct?

2. There is a better approach to try to prove the statement of the title? Thank you in advance.

Answer 1

You can define a bijection $N\times G/N\to G$ in the following way.

For each $[g]\in G/N$, choose an element $v([g])\in[g]$. Define $$ \alpha\colon N\times G/N\to G, \qquad \alpha(n,[g])=v([g])n $$ Note that $v([g])N=[g]$.

Suppose $\alpha(n_1,[g_1])=\alpha(n_2,[g_2])$. Then, since $v([g_2])N=[g_2]$, by definition, we have $v([g_1])n_1\in [g_2]$, so $[g_1]=[g_2]$ and therefore $v([g_1])=v([g_2])$, hence also $n_1=n_2$.

The map is also surjective: let $g\in G$; then $v([g])\in gN$ and so $$ n=g^{-1}v([g])\in N $$ so that $$ g=v([g])n^{-1}=\alpha(n^{-1},[g])= \alpha(,v([g])^{-1}g,[g]) $$

Your idea seems similar to this one, but the proof is incorrect. You wanted to define the inverse of my $\alpha$, but it's not $N$ that you want to “enumerate”; rather the “enumeration” is made of $G/N$ via the map $v$.

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Is there a better approach? Yes. First observe that, for each $g\in G$, the map $N\to gN$ defined by $n\mapsto gn$ is bijective. Since the cosets form a partition of $G$ and each element in the partition has the same cardinality as $N$, we get that $|G|$ is the product of $N$ by the number of elements of $G/N$, that is, the number of distinct cosets.