proof of fourth isomorphism theorem
First we must prove that the map defined by $A\mapsto A/N$ is a bijection. Let $\theta $ denote this map, so that $\theta (A)=A/N$. Suppose $A/N=B/N$, then for any $a\in A$ we have $aN=bN$ for some $b\in B$, and so ${b}^{1}a\in N\subseteq B$. Hence $A\subseteq B$, and similarly $B\subseteq A$, so $A=B$ and $\theta $ is injective^{}. Now suppose $S$ is a subgroup^{} of $G/N$ and $\varphi :G\to G/N$ by $\varphi (g)=gN$. Then ${\varphi}^{1}(S)=\{s\in G:sN\in S\}$ is a subgroup of $G$ containing $N$ and $\theta ({\varphi}^{1}(S))=\{sN:sN\in S\}=S$, proving that $\theta $ is bijective^{}.
Now we move to the given properties:

1.
$A\le B$ iff $A/N\le B/N$
If $A\le B$ then trivially $A/N\le B/N$, and the converse^{} follows from the fact that $\theta $ is bijective.

2.
$A\le B$ implies $B:A=B/N:A/N$
Let $\psi $ map the cosets in $B/A$ to the cosets in $(B/N)/(A/N)$ by mapping the coset $bA$ $b\in B$ to the coset $(bN)(A/N)$. Then $\psi $ is well defined and injective because:
${b}_{1}A={b}_{2}A$ $\iff {b}_{1}^{1}{b}_{2}\in A$ $\iff {({b}_{1}N)}^{1}({b}_{2}N)={b}_{1}^{1}{b}_{2}N\in A/N$ $\iff ({b}_{1}N)(A/N)=({b}_{2}N)(A/N).$ Finally, $\psi $ is surjective since $b$ ranges over all of $B$ in $(bN)(A/N)$.

3.
$\u27e8A,B\u27e9/N=\u27e8A/N,B/N\u27e9$
To show $\u27e8A,B\u27e9/N\subseteq \u27e8A/N,B/N\u27e9$ we need only show that if $x\in A$ or $x\in B$ then $xN\in \u27e8A/N,B/N\u27e9$. The other cases are dealt with using the fact that $(xy)N=(xN)(yN)$. So suppose $x\in A$ then clearly $xN\in \u27e8A/N,B/N\u27e9$ because $xN\in A/N$. Similarly for $x\in B$. Similarly, to show $\u27e8A/N,B/N\u27e9\subseteq \u27e8A,B\u27e9/N$ we need only show that if $xN\in A/N$ or $xN\in B/N$ then $x\in \u27e8A,B\u27e9$. So suppose $xN\in A/N$, then $xN=aN$ for some $a\in A$, giving ${a}^{1}x\in N\subseteq A$ and so $x\in A\subseteq \u27e8A,B\u27e9$. Similarly for $xN\in B/N$.

4.
$(A\cap B)/N=(A/N)\cap (B/N)$
Suppose $xN\in (A\cap B)/N$, then $xN=yN$ for some $y\in (A\cap B)$ and since $N\subseteq (A\cap B)$, $x\in (A\cap B)$. Therefore $x\in A$ and $x\in B$, and so $xN\in (A/N)\cap (B/N)$ meaning $(A\cap B)/N\subseteq (A/N)\cap (B/N)$. Now suppose $xN\in (A/N)\cap (B/N)$. Then $xN=aN$ for some $a\in A$, giving ${a}^{1}x\in N\subseteq A$ and so $x\in A$. Similarly $x\in B$, therefore $xN\in (A\cap B)/N$ and $(A/N)\cap (B/N)\subseteq (A\cap B)/N$.

5.
$A\u22b4G$ iff $(A/N)\u22b4(G/N)$
Suppose $A\u22b4G$. Then for any $g\in G$ we have $(gN)(A/N){(gN)}^{1}=(gA{g}^{1})/N=A/N$ and so $(A/N)\u22b4(G/N)$.
Conversely suppose $(A/N)\u22b4(G/N)$. Consider $\sigma :g\mapsto (gN)(A/N)$, the composition^{} of the map from $G$ onto $G/N$ and the map from $G/N$ onto $(G/N)/(A/N)$. $g\in \mathrm{ker}\pi $ iff $(gN)(A/N)=(A/N)$ which occurs iff $gN\in A/N$ therefore $gN=aN$ for some $a\in A$. However $N$ is contained in $A$, so this statement is equivalnet to saying $g\in A$. So $A$ is the kernel of a homomorphism^{}, hence is a normal subgroup^{} of $G$.
Title  proof of fourth isomorphism theorem 

Canonical name  ProofOfFourthIsomorphismTheorem 
Date of creation  20130322 14:17:38 
Last modified on  20130322 14:17:38 
Owner  aoh45 (5079) 
Last modified by  aoh45 (5079) 
Numerical id  9 
Author  aoh45 (5079) 
Entry type  Proof 
Classification  msc 20A05 