Let \(H\) be a subgroup of a group missing\(G\). A left coset of \(H\) in \(G\) is a subset of \(G\) that is of the form \(xH\), where \(x\in G\) and \(xH=\{xh:h\in H\}\). Similarly a right coset of \(H\) in \(G\) is a subset of \(G\) that is of the form \(Hx\), where \(Hx=\{hx:h\in H\}\)
Note that a subgroup \(H\) of a group \(G\) is itself a left coset of \(H\) in \(G\).
Let \(H\) be a subgroup of a group \(G\), and let \(x\) and \(y\) be elements of \(G\). Suppose that \(xH\cap yH\) is non-empty. Then \(xH=yH\).
Let \(z\) be some element of \(xH\cap yH\). Then \(z=xa\) for some \(a\in H\), and \(z=yb\) for some \(b\in H\). If \(h\) is any element of \(H\) then \(ah\in H\) and \(a^{-1}h\in H\), since \(H\) is a subgroup of \(G\). But \(zh=x(ah)\) and \(xh=z(a^{-1}h)\) for all \(h\in H\). Therefore \(zH\subset xH\) and \(xH\subset zH\), and thus \(xH=zH\). Similarly \(yH=zH\), and thus \(xH=yH\), as required.∎∎
Let \(H\) be a finite subgroup of a group \(G\). Then each left coset of \(H\) in \(G\) has the same number of elements as \(H\).
Let \(H=\{h_{1},h_{2},\ldots,h_{m}\}\), where \(h_{1},h_{2},\ldots,h_{m}\) are distinct, and let \(x\) be an element of \(G\). Then the left coset \(xH\) consists of the elements \(xh_{j}\) for \(j=1,2,\ldots,m\). Suppose that \(j\) and \(k\) are integers between \(1\) and \(m\) for which \(xh_{j}=xh_{k}\). Then \(h_{j}=x^{-1}(xh_{j})=x^{-1}(xh_{k})=h_{k}\), and thus \(j=k\), since \(h_{1},h_{2},\ldots,h_{m}\) are distinct. It follows that the elements \(xh_{1},xh_{2},\ldots,xh_{m}\) are distinct. We conclude that the subgroup \(H\) and the left coset \(xH\) both have \(m\) elements, as required.∎∎
(Lagrange’s Theorem)\label{Lagrange} Let \(G\) be a finite group, and let \(H\) be a subgroup of \(G\). Then the order of \(H\) divides the order of \(G\).
Each element \(x\) of \(G\) belongs to at least one left coset of \(H\) in \(G\) (namely the coset \(xH\)), and no element can belong to two distinct left cosets of \(H\) in \(G\) (see Lemma \ref{LeftCosetsDisjoint}). Therefore every element of \(G\) belongs to exactly one left coset of \(H\). Moreover each left coset of \(H\) contains \(|H|\) elements (Lemma \ref{SizeOfLeftCoset}). Therefore \(|G|=n|H|\), where \(n\) is the number of left cosets of \(H\) in \(G\). The result follows.∎∎
I am going to add a citation to Emily (Burbelo 2010) (Burbelo 2010a)
I am going to add a citation to Miki’s sister (Agrawal 2004)
One more Pubmed (Jacob ). Test.
Website here: (One 1999)