Proof of Left Cosets Partition a Group and All Have the Same Cardinality

Throughout, associativity refers to condition 1 of Group and Abelian Group, and each use of it is applied to an explicitly named triple of elements of $G$. We use the subgroup conditions from Subgroup and the identities of Cancellation Laws and Basic Inverse Identities in a Group.

Claim 1. By condition 1 of Subgroup we have $e_G\in H$, and $ae_G=a$ by condition 2 of Group and Abelian Group. Hence $a\in aH$. Taking $a=e_G$ gives $e_GH=\{e_Gh: h\in H\}=H$.

Claim 2. Suppose first that $aH=bH$. By Claim 1 we have $b\in bH=aH$, so $b=ah$ for some $h\in H$. Then, by associativity applied to the triple $a^{-1},a,h$,

Conversely, suppose $h_0=a^{-1}b\in H$. By associativity applied to the triple $a,a^{-1},b$,

Now let $h\in H$. Then $h_0h\in H$ by condition 2 of Subgroup, and associativity applied to the triple $a,h_0,h$ gives

Hence $bH\subseteq aH$.

For the reverse inclusion, observe that by claims 3 and 2 of Cancellation Laws and Basic Inverse Identities in a Group,

so $b^{-1}a=h_0^{-1}$, which lies in $H$ by condition 3 of Subgroup. Applying the argument of the previous paragraph with the roles of $a$ and $b$ exchanged, and with $h_0^{-1}$ in place of $h_0$, yields $aH\subseteq bH$. Therefore $aH=bH$.

Claim 3. Suppose $c\in aH\cap bH$. Since $c\in aH$, we have $c=ah$ for some $h\in H$, and as in the first paragraph of the proof of Claim 2 this gives $a^{-1}c=h\in H$; by Claim 2 applied to the pair $a,c$ we get $aH=cH$. Since $c\in bH$, the same argument applied to the pair $b,c$ gives $bH=cH$. Hence $aH=bH$.

Combining this with Claim 1: every $x\in G$ lies in the left coset $xH$, and if $x$ lies in two left cosets $aH$ and $bH$, then $aH\cap bH\ne\emptyset$, so $aH=bH$. Thus $x$ lies in exactly one left coset.

Claim 4. Let $a\in G$ and define $\lambda(h)=ah$ for $h\in H$. By definition of $aH$ the map $\lambda$ takes values in $aH$, and every element of $aH$ is of the form $ah$ with $h\in H$, so $\lambda$ is surjective onto $aH$. If $\lambda(h)=\lambda(h')$, then $ah=ah'$, and the left cancellation law, claim 1 of Cancellation Laws and Basic Inverse Identities in a Group, gives $h=h'$. Hence every element of $aH$ has exactly one preimage in $H$, so $\lambda$ is a bijection from $H$ onto $aH$.

Finally, suppose $H$ has $t$ elements, and let $f:[t]\to H$ be a bijection from the initial segment $[t]$ onto $H$, which exists by Number of Elements of a Set. By claim 2 of Injectivity, Composition, and Restriction of Bijections, the composite $i\mapsto\lambda(f(i))$ is a bijection from $[t]$ onto $aH$. Hence $aH$ has $t$ elements.