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 $\phi:G\to G/N$ by $\phi(g)=gN$ . Then $\phi^{-1}(S)=\{s\in G:sN\in S\}$ is a subgroup of $G$ containing $N$ and $\theta(\phi^{-1}(S))=\{sN:sN\in S\}=S$ , proving that $\theta$ is bijective.

Now we move to the given properties:

1.

$A\leq B$ iff $A/N\leq B/N$

If $A\leq B$ then trivially $A/N\leq B/N$ , and the converse follows from the fact that $\theta$ is bijective.
2.

$A\leq B$ implies $\left|B:A\right|=\left|B/N:A/N\right|$

Let $\psi$ map the cosets in $B/A$ to the cosets in $(B/N)/(A/N)$ by mapping the coset $b A$ $b\in B$ to the coset $(bN)(A/N)$ . Then $\psi$ is well defined and injective because:

$\displaystyle b_{1}A=b_{2}A$ $\displaystyle\iff b_{1}^{-1}b_{2}\in A$

$\displaystyle\iff(b_{1}N)^{-1}(b_{2}N)=b_{1}^{-1}b_{2}N\in A/N$

$\displaystyle\iff(b_{1}N)(A/N)=(b_{2}N)(A/N).$

Finally, $\psi$ is surjective since $b$ ranges over all of $B$ in $(bN)(A/N)$ .
3.

$\langle{A,B}\rangle/N=\langle{A/N,B/N}\rangle$

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

$A\unlhd G$ iff $(A/N)\unlhd(G/N)$

Suppose $A\unlhd G$ . Then for any $g\in G$ we have $(gN)(A/N)(gN)^{-1}=(gAg^{-1})/N=A/N$ and so $(A/N)\unlhd(G/N)$ .
Conversely suppose $(A/N)\unlhd(G/N)$ . Consider $\sigma\colon g\mapsto(gN)(A/N)$ , 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 $G$ .

Title	proof of fourth isomorphism theorem
Canonical name	ProofOfFourthIsomorphismTheorem
Date of creation	2013-03-22 14:17:38
Last modified on	2013-03-22 14:17:38
Owner	aoh45 (5079)
Last modified by	aoh45 (5079)
Numerical id	9
Author	aoh45 (5079)
Entry type	Proof
Classification	msc 20A05

	$\displaystyle b_{1}A=b_{2}A$	$\displaystyle\iff b_{1}^{-1}b_{2}\in A$
		$\displaystyle\iff(b_{1}N)^{-1}(b_{2}N)=b_{1}^{-1}b_{2}N\in A/N$
		$\displaystyle\iff(b_{1}N)(A/N)=(b_{2}N)(A/N).$