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. $⟨A,B⟩/N=⟨A/N,B/N⟩$

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

   Suppose $A⊴G$. Then for any $g\in G$ we have $(gN)(A/N){(gN)}^{-1}=(gA{g}^{-1})/N=A/N$ and so $(A/N)⊴(G/N)$.
   Conversely suppose $(A/N)⊴(G/N)$. Consider $\sigma :g\mapsto (gN)(A/N)$, the composition of the map from $G$ onto $G/N$ and the map from $G/N$ onto $(G/N)/(A/N)$. $g\in \ker \pi$ 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 $G$.