Kernels normality and quotient structure

A quotient can compress a group while preserving its operation. The compression is useful only if multiplication remains independent of representatives. We derive the exact compatibility condition, identify quotients arising from homomorphisms, and prove a correspondence between subgroups before and after compression.

1 Normal subgroups as compatible equivalence classes

For $N\le G$, normality means $gNg^{-1}=N$ for every $g\in G$. This is equivalent to $gN=Ng$: multiply the first equality on the right by $g$, or the second on the right by $g^{-1}$. In an abelian group every subgroup is normal. For nonabelian groups it must be checked.

The kernel of a homomorphism $\phi:G\to K$ is normal. If $n$ lies in the kernel, then $\phi(gng^{-1})=\phi(g)e\phi(g)^{-1}=e$, giving one inclusion; replacing $g$ by $g^{-1}$ gives the other. Notice the distinction between an inclusion proof and the equality required by normality.

Theorem. The rule $(aN)(bN)=abN$ is well defined on left cosets if and only if $N$ is normal.

Proof. For normal $N$, replace representatives by $an_1$ and $bn_2$. Their product is $ab(b^{-1}n_1b)n_2$, which lies in $abN$. Thus any choices give the same output coset. Conversely, if the rule is well defined, $nN=N$ implies $ngN=gN$ for every $n\in N$, so $g^{-1}ng\in N$. This gives $g^{-1}Ng\subseteq N$ for every $g$, and the reverse inclusion follows by conjugating the corresponding statement for $g^{-1}$. $\square$

The quotient group $G/N$ has identity $N$ and inverse $(aN)^{-1}=a^{-1}N$. Associativity is inherited from $G$. The projection $\pi:G\to G/N$, $a\mapsto aN$, is onto with kernel $N$. Hence normal subgroups are precisely kernels of group homomorphisms.

2 The first isomorphism theorem as a complete proof

For $\phi:G\to K$, define $\bar\phi:G/\ker\phi\to\operatorname{im}\phi$ by $\bar\phi(g\ker\phi)=\phi(g)$.

To prove it is well defined, suppose $g\ker\phi=h\ker\phi$. Then $g^{-1}h\in\ker\phi$, so $\phi(g)^{-1}\phi(h)=e$. Conversely, equality of the two images gives this kernel condition, proving injectivity as well. Every element of the image has a representative $\phi(g)$, proving surjectivity. Finally,

$$\bar\phi((g\ker\phi)(h\ker\phi))=\phi(gh)=\phi(g)\phi(h).$$

Thus

$$G/\ker\phi\cong\operatorname{im}\phi.$$

Each step is necessary: naming a formula alone does not prove it defines an isomorphism.

Worked determinant example. The determinant map $GL_2(\mathbb R)\to\mathbb R^\times$ is a homomorphism because determinants multiply. It is onto: $\operatorname{diag}(a,1)$ has determinant $a$ for every nonzero real $a$. Its kernel is $SL_2(\mathbb R)$, the determinant-one matrices. Therefore

$$GL_2(\mathbb R)/SL_2(\mathbb R)\cong\mathbb R^\times.$$

This does not say the quotient retains every matrix entry. It retains exactly the multiplicative determinant information.

3 Subgroup correspondence

Correspondence theorem. For $N\trianglelefteq G$, subgroups of $G/N$ correspond bijectively to subgroups $H\le G$ containing $N$, by $H\mapsto H/N$ and $L\mapsto\pi^{-1}(L)$. Normal subgroups correspond to normal subgroups.

Proof. A preimage of a subgroup is a subgroup by the homomorphism laws and contains the projection kernel. If $H$ contains $N$, then the preimage of $H/N$ is $H$: membership of $gN$ means $gN=hN$ for some $h\in H$, giving $g=hn\in H$. Conversely, surjectivity of $\pi$ gives $\pi(\pi^{-1}(L))=L$. For normality, conjugating $hN$ by $gN$ gives $(ghg^{-1})N$; this lies in $H/N$ for every $g$ exactly when $H$ is normal, using $N\subseteq H$. $\square$

If additionally $N\subseteq H\trianglelefteq G$, the map $G/N\to G/H$, $gN\mapsto gH$, has kernel $H/N$. The first theorem gives the third isomorphism theorem:

$$(G/N)/(H/N)\cong G/H.$$

The second isomorphism theorem follows by restricting $\pi$ to any subgroup $K\le G$: kernel $K\cap N$, image $KN/N$, hence $K/(K\cap N)\cong KN/N$. Normality of $N$ ensures $KN$ is a subgroup, since products can move an $N$ factor across a $K$ factor by conjugation.

For the supplement below, the second isomorphism argument resets notation to an arbitrary subgroup $H\le G$ and a normal subgroup $N\trianglelefteq G$. In its third isomorphism argument the assumptions are instead $N\subseteq K\trianglelefteq G$ with $N\trianglelefteq G$. In particular $G/K$ is a group in that argument; no such normality is asserted for an arbitrary subgroup used earlier.

Representative failure and the full projection-map arguments

Take $H=\{e,(1\ 2)\}$ in $S_3$ and $b=(1\ 3)$. The same coset $H$ has representatives $e$ and $(1\ 2)$. Using $e$ in its product with $bH$ gives $bH$; using $(1\ 2)$ gives $(1\ 2)bH$, which is different because $b^{-1}(1\ 2)b=(2\ 3)\notin H$. Thus a rule on representatives can fail to define any operation on cosets at all. Merely counting the cosets cannot repair that failure.

Here are the closure details for the second theorem. The set $HN$ is nonempty because it contains $e$. For $h_1n_1,h_2n_2\in HN$,

$$(h_1n_1)(h_2n_2)^{-1} =h_1h_2^{-1}\bigl(h_2n_1n_2^{-1}h_2^{-1}\bigr)\in HN.$$

The first factor is in $H$ and the parenthesized factor is in $N$ by normality. The subgroup test proves $HN\le G$. Normality of $N$ in $G$ implies normality in $HN$. For the restricted projection $h\mapsto hN$, the kernel is exactly $H\cap N$; every coset of an element $hn\in HN$ equals $hN$, so its image is all of $HN/N$. These statements justify every group and every map used in the theorem.

For the third theorem, $gN=hN$ gives $h^{-1}g\in N\subseteq K$, hence $gK=hK$. Therefore $gN\mapsto gK$ is well defined. It preserves products since $(gh)K=(gK)(hK)$, and is onto because each $gK$ is the image of $gN$. Its kernel consists precisely of the cosets $kN$ with $k\in K$, which form $K/N$. A kernel is normal, so $K/N\trianglelefteq G/N$ follows before taking the final quotient.

4 Graded practice

E1 Foundation. Prove a subgroup of index 2 is normal without enumerating a multiplication table.

E2 Cyclic maps. Classify homomorphisms $\mathbb Z/6\mathbb Z\to\mathbb Z/9\mathbb Z$. Find each kernel and image.

E3 Correspondence. List the subgroups of $\mathbb Z/12\mathbb Z$ by lifting them to subgroups of $\mathbb Z$ containing $12\mathbb Z$.

E4 Quotient in stages. Identify $(\mathbb Z/18\mathbb Z)/(6\mathbb Z/18\mathbb Z)$ and describe its natural map explicitly.

E5 Nonexample. In $S_3$, show $\langle(1\ 2)\rangle$ is not normal and explain which step of quotient multiplication fails.

5 Full solutions

S1. If $g\in H$, both cosets $gH$ and $Hg$ equal $H$. Otherwise each is the complement of $H$, because the index-2 left and right partitions each have only one other block. Thus $gH=Hg$ for every $g$.

S2. A map is determined by $a=\phi([1])$, and the relation $6[1]=0$ requires $6a=0$ modulo 9. Thus $a=0,3,6$. For $a=0$, kernel is the whole domain and image is zero. For $a=3$ or 6, the kernel is $\{[0],[3]\}$ and image is $\{[0],[3],[6]\}$. The two nonzero maps have quotient domain of order 3, agreeing with their images.

S3. Every subgroup of $\mathbb Z$ is $d\mathbb Z$. Containment $12\mathbb Z\subseteq d\mathbb Z$ means $d\mid12$. Thus $d=1,2,3,4,6,12$ gives all six quotient subgroups, with orders $12/d$. The correspondence proves completeness of the list.

S4. The map sends $[a]_{18}$ to $[a]_6$ and has kernel $\{[0]_{18},[6]_{18},[12]_{18}\}$. The quotient is $\mathbb Z/6\mathbb Z$ by the first or third theorem.

S5. Conjugating $(1\ 2)$ by $(1\ 3)$ gives $(3\ 2)$ outside the subgroup. In the necessity proof, replacing $eN$ by $nN$ can change the product with $gN$, because $g^{-1}ng$ need not lie in $N$. Thus the proposed multiplication depends on representatives.

Summary

Normality is exactly the condition for group quotient multiplication. The first isomorphism theorem identifies a map's retained information, and the correspondence theorem tracks subgroup structure through the projection. Further isomorphism results follow from explicit maps and their kernels rather than from symbolic cancellation of quotient notation.