# Proof

(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 $(G/N,\circ ).$

• Ad $$(5)$$: $$(G/N,\circ )$$ is Abelian, if $G$ is.
• This follows immediately from the definition of $"\circ".$

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013