Editing Quotient group (section)

=== Even and odd integers ===
Consider the group of [[integer]]s <math>\Z</math> (under addition) and the subgroup <math>2\Z</math> consisting of all even integers. This is a normal subgroup, because <math>\Z</math> is [[abelian group|abelian]]. There are only two cosets: the set of even integers and the set of odd integers, and therefore the quotient group <math>\Z\,/\,2\Z</math> is the cyclic group with two elements. This quotient group is isomorphic with the set <math>\left\{0,1 \right\}</math> with addition modulo&nbsp;2; informally, it is sometimes said that <math>\Z\,/\,2\Z</math> ''equals'' the set <math>\left\{0,1 \right\}</math> with addition modulo 2.

'''Example further explained...'''
: Let <math> \gamma(m) </math> be the remainders of <math> m \in \Z </math> when dividing by {{tmath|1= 2 }}. Then, <math> \gamma(m)=0 </math> when <math> m </math> is even and <math> \gamma(m)=1 </math> when <math> m </math> is odd.
: By definition of {{tmath|1= \gamma }},  the kernel of {{tmath|1= \gamma }}, {{tmath|1= \ker(\gamma) = \{ m \in \Z : \gamma(m)=0 \} }},  is the set of all even integers.
: Let {{tmath|1= H = \ker(\gamma) }}. Then, <math> H </math> is a subgroup, because the identity in {{tmath|1= \Z }}, which is {{tmath|1= 0 }}, is in {{tmath|1= H }}, the sum of two even integers is even and hence if <math> m </math> and <math> n </math> are in {{tmath|1= H }}, <math> m+n </math> is in <math> H </math> (closure) and if <math> m </math> is even, <math> -m </math> is also even and so <math> H </math> contains its inverses.
: Define <math> \mu : \mathbb{Z} / H \to \mathrm{Z}_2 </math> as <math> \mu(aH)=\gamma(a) </math> for <math> a\in\Z </math> and <math>\mathbb{Z} / H</math> is the quotient group of left cosets; {{tmath|1= \mathbb{Z} / H=\{H,1+H\} }}.
: Note that we have defined {{tmath|1= \mu }}, <math> \mu(aH) </math> is <math> 1 </math> if <math> a </math> is odd and <math> 0 </math> if <math> a </math> is even.
: Thus, <math> \mu </math> is an isomorphism from <math>\mathbb{Z} / H</math> to {{tmath|1= \mathrm{Z}_2 }}.