fourth isomorphism theorem
Theorem 1 (The Fourth Isomorphism Theorem)
Let $G$ be a group and $N\mathrm{\u22b4}G$. There is a bijection between $G\mathit{}\mathrm{(}N\mathrm{)}$, the set of subgroups^{} of $G$ containing $N$, and the set of subgroups of $G\mathrm{/}N$ defined by $A\mathrm{\to}A\mathrm{/}N$. Moreover, for any two subgroups $A\mathrm{,}B$ in $G\mathit{}\mathrm{(}N\mathrm{)}$, we have

1.
$A\le B$ if and only if $A/N\le B/N$,

2.
$A\le B$ implies $B:A=B/N:A/N$,

3.
$\u27e8A,B\u27e9/N=\u27e8A/N,B/N\u27e9$,

4.
$(A\cap B)/N=(A/N)\cap (B/N)$, and

5.
$A\u22b4G$ if and only if $(A/N)\u22b4(G/N)$.
Title  fourth isomorphism theorem 

Canonical name  FourthIsomorphismTheorem 
Date of creation  20130322 14:00:52 
Last modified on  20130322 14:00:52 
Owner  bwebste (988) 
Last modified by  bwebste (988) 
Numerical id  16 
Author  bwebste (988) 
Entry type  Theorem 
Classification  msc 20A05 
Synonym  correspondence theorem 
Synonym  lattice isomorphism theorem 