(related to Lemma: Factor Groups)
Let \( ( G ,\ast) \) be a group, \(N\unlhd G\) its normal subgroup and the operation $"\circ":=(a_1 N)\circ(a_2N):=(a_1\ast a_2)N$ for all $a_1,a_2\in G$ be given. * Ad \((1)\): The operation "\(\circ\)" is well-defined, (i.e. it does not depend on the particular choice of representatives \(a_1,a_2\)). * Let \(a_1,b_1\) be two different representatives of the left coset \(a_1N\) and \(a_2,b_2\) be two different representatives of the left coset \(a_2N\). * If \(b_1\in a_1N\) then there exists an \(c_1\in N\) with \(b_1=a_1\ast c_1\). * Analogously, there is an \(c_2\in N\) with \(b_2=a_2\ast c_2\). * Therefore (using the associativity of the operation "\(\ast\)"): \[\begin{array}{ccl} (b_1\ast b_2)N&=&((a_1\ast c_1)\ast (a_2\ast c_2))N\\ &=&(a_1\ast c_1\ast a_2)\ast\underbrace{(c_2N)}_{=N\text{ because }c_2\in N}\\ &=&(a_1\ast (c_1\ast a_2))N \end{array}\] * Because \(N\) is a normal subgroup of \(G\), there is a \(c_3\in N\) with \(c_1\ast a_2=a_2\ast c_3\). * Therefore, continue the calculation by replacing \[\begin{array}{ccl} (a_1\ast (c_1\ast a_2))N&=&(a_1\ast (a_2\ast c_3))N\\ &=&(a_1\ast a_2)\ast\underbrace{(c_3N)}_{=N\text{ because }c_3\in N}\\ &=&(a_1\ast a_2)N \end{array}\] * This demonstrates that $(a_1\ast a_2)N=(b_1\ast b_2)N.$ * Ad \((2)\) The operation "\(\circ\)" is associative. * This follows from the associativity of "\(\ast\)", since for all \(a_1,a_2,a_3\in G\): \[\begin{array}{ccl} ((a_1 N)\circ(a_2N))\circ(a_3N)&=&(a_1\ast a_2)N\circ (a_3N)\\ &=&((a_1\ast a_2)\ast a_3)N\\ &=&(a_1\ast (a_2\ast a_3))N\\ &=&(a_1N)\circ(a_2\ast a_3)N\\ &=&(a_1N)\circ((a_2N\circ a_3N))\\ \end{array}\] * Ad \((3)\): The factor group \((G/N,\circ)\) contains at least one element \(N\) and this is its neutral element. * The left coset \(e_GN\) with respect to the neutral element \(e_G\in G\) is an element of \((G/N,\circ)\). * Thus, \(N=e_GN\in G/N\). * Moreover, for any coset \(aN\in G/N\) we have \[\begin{array}{ccccccc} N\circ(aN)&=&(e_GN)\circ(aN)&=&(e_G\ast a)N&=&aN\\ (aN)\circ N&=&(aN)\circ(e_GN)&=&(a\ast e_G)N&=&aN\\ \end{array}\] * Therefore, \(N\) is the identity element of \((G/N,\circ)\). * Ad \((4)\): The inverse element of an element \(aN\) of \((G/N,\circ)\) is \(a^{-1}N\). * For any \(aN\in G/N\) we have \[\begin{array}{ccccccc} (a^{-1}N)\circ(aN)&=&(a^{-1}\ast a) N&=&e_GN&=&N\\ (aN)\circ(a^{-1}N)&=&(a\ast a^{-1}) N&=&e_GN&=&N.\\ \end{array}\]
Altogether, we have shown that the quotient set \(G/N:=\{aN: a\in G\}\) of all (left) cosets, together with the binary operation "$\circ$" forms a factor group $(G/N,\circ ).$
