A question regarding the proof of G a finite group, H a subgroup then $\left|G\right| = \left|G/H\right|\cdot \left|H\right| $

Question

I'm currently looking at the proof of the following statement:

let $\left(G,\cdot \right)$ be a finite group, $H$ a subgroup of $G$, then:$$\left|G\right| = \left|G/H\right|\cdot \left|H\right| $$

Now apparently it is sufficient to show that the number of elements in the left coset $a\cdot H$ is equal to the number of elements in the subgroup $H$ and that is done by proving that $\varphi:H \to a\cdot H$; $\varphi\left(h\right)=a\cdot h $ is a bijection. I understand why is that a bijection and why is the number of elements in the left coset equal to the number of elements in the subgroup, but how does that prove the statement? What am I missing here?

Answer 1

Because $G$ is the disjoint union of the conjugate clases (the sets $a.H$) and there are $G/H$ such classes.

Answer 2

Since all cosets have the same cardinality you can sort the elements of $G$ into a rectangular lattice, where the length of one side is $|H|$ (the size of any coset) and the length of the other side is $|G/H|$ (the amount of cosets).

Answer 3

That

$\vert aH \vert = \vert H \vert, \tag 1$

i.e. that the cardinality of any coset is the same as the cardinality of the subgroup, is only half the story. The other half is:

(C) Cosets $aH$ of $H$ are either disjoint or identical.

We may prove this as follows: if

$aH \cap bH \ne \emptyset, \tag 2$

then there are

$h_1, h_2 \in H \tag 3$

with

$ah_1 = bh_2; \tag 4$

then

$a = bh_2h_1^{-1} \in bH, \tag 5$

and similarly

$b \in aH; \tag 6$

(5) and (6) are sufficient to force

$aH = bH, \tag 7$

since they show

$bH \subseteq aH, \; aH \subseteq bH; \tag 8$

thus we see that (C) must bind, which means that the distinct cosets of $H$ form a partition of $G$ into disjoint subsets of equal cardinality (and we note here that every $g \in G$ is in some coset, viz. $g \in gH$); the number of such cosets is

$[G:H] = \vert G \vert / \vert H \vert. \tag 9$

Answer 4

Consider the following relation on the set $G$: $$ a\sim_H b \qquad\text{if and only if}\qquad a^{-1}b\in H $$ This is an equivalence relation:

1. (reflexivity) $a\sim_H a$, because $a^{-1}a=1\in H$;
2. (symmetry) if $a\sim_H b$, then $a^{-1}b\in H$; therefore $(a^{-1}b)^{-1}=b^{-1}a\in H$, and so $b\sim_H a$;
3. (transitivity) if $a\sim_H b$ and $b\sim_H c$, then $a^{-1}b\in H$ and $b^{-1}c\in H$; therefore $(a^{-1}b)(b^{-1}c)=a^{-1}c\in H$ and so $a\sim_H c$.

Every equivalence relation on a set defines a partition of the same set via the equivalence classes: if $[a]_{\sim_H}=\{x\in G: a\sim_H x\}$, then two sets $[a]_{\sim_H}$ and $[b]_{\sim_H}$ are either equal or disjoint. It also holds, by reflexivity, that $a\in[a]_{\sim_H}$.

Now we can prove that $[a]_{\sim_H}=aH$. Indeed, if $a\sim_H x$, then $a^{-1}x\in H$, so $x=a(a^{-1}x)\in aH$; conversely, if $h\in H$, then $a\sim_H(ah)$, because $a^{-1}(ah)=h\in H$.

Since the equivalence classes are a partition of $G$, in order to count the elements of $G$ we just need to count the number of elements in each equivalence class and sum them up. However, you have proved that two equivalence classes (that is, the left cosets) have the same number of elements. Therefore $|G|$ is the number of elements in one coset times the number of cosets: $$ |G|=|H|\cdot|G/H| $$ because $H=1H$ is indeed one of the cosets.

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If you consider the relation $\sim_H'$ defined by $$ a\sim_H' b \qquad\text{if and only if}\qquad ab^{-1}\in H $$ then the verifications are essentially the same as before, with the difference that $[a]_{\sim_H'}=Ha$, the right coset. Also the right cosets have the same cardinality as $H$, so we have proved that there are as many left cosets as right cosets.

This can also be proved by noticing that the map $$ aH\mapsto Ha^{-1} $$ is a bijection of the set of left cosets onto the set of right cosets (this doesn't require finiteness of the group $G$, so it's interesting on its own).