(related to Proposition: Properties of Cosets)
In the following let \((G,\ast)\) be a group and \(H\subseteq G\) its subgroup.
If \(|H|=n\), then \(|Ha|\le n\), which follows from the definition of a coset. But \(|Ha| \not < n\). For if \(|Ha| < n\), then there would be at least two different elements \(g,h\in H\), \(g\neq h\) and \(g\ast a=h\ast a\). This is equivalent to \(g=h\), which is a contradiction to \(g\neq h\). Thus it must be \(|Ha|=|H|\). Analogously, we can prove \(|aH|=|H|\).
We have to prove that \(a\sim_l b:=a\in bH\) (respectively \(a\sim_r b:=a\in Hb\)) define equivalence relations, by proving their reflexivity, symmetry and transitivity.
Because \(H\subset G\) is a subgroup of \(G\), the neutral element of \(G\) is also the neutral element of \(H\). Therefore, following the definition of cosets, \(a\in aH\) and \(a\in Ha\). So \(a\sim_l a\) (respectively \(a\sim_r a\)) are reflexive.
Assume \(a\sim_l b\). From \(a\in bH\) it follows that there exists \(x\in H\) with \(a=b\ast x\), which is equivalent to \(a\ast x^{-1}=b\). Because also \(x^{-1}\in H\), it follows that \(b\in aH\), so \(b\sim_l a\) is symmetric. Respectively, assume \(a\sim_r b\). From \(a\in Hb\) it follows that there exists \(x\in H\) with \(a=x\ast b\), which is equivalent to \(x^{-1}a=b\). Because also \(x^{-1}\in H\), it follows that \(b\in Ha\), so \(b\sim_r a\) is symmetric.
Assume \(a\sim_l b\) and \(b\sim_l c\). From \(a\in bH\) it follows that there exists \(x\in H\) with \(a=b\ast x\), which is equivalent to \(a\ast x^{-1}=b\). From \(b\in cH\) it follows that there exists \(y\in H\) with \(b=c\ast y\). From both results, it follows that \(a\ast x^{-1}=c\ast y\), which is equivalent to \(a=c\ast y\ast x\). Because \(H\) is closed under \(\ast\), \(y\ast x\in H\). Therefore, \(a\sim_l c\) is transitive. Respectively, assume \(a\sim_r b\) and \(b\sim_r c\). From \(a\in Hb\) it follows that there exists \(x\in H\) with \(a=x\ast b\), which is equivalent to \(x^{-1}\ast a=b\). From \(b\in Hc\) it follows that there exists \(y\in H\) with \(b=y\ast c\). From both results, it follows that \(x^{-1}\ast a=y\ast c\), which is equivalent to \(a=x\ast y\ast c\). Because \(H\) is closed under \(\ast\), \(x\ast y\in H\). Therefore, \(a\sim_r c\) is transitive.
