Normal subgroups and isomorphism theorems

A quotient deliberately forgets distinctions. For its multiplication to make sense, the answer cannot depend on which representative we choose. Normality is exactly the compatibility condition for this forgetting process in a group. Homomorphisms then explain what information a map forgets and what structure remains.

1 Homomorphisms and kernels

A group homomorphism $\phi:G\to K$ satisfies $\phi(ab)=\phi(a)\phi(b)$. It sends the identity to the identity, since $\phi(e)=\phi(e)^2$ and cancellation applies; it sends inverses to inverses by applying it to $aa^{-1}=e$. Its kernel is $\ker\phi=\{g:\phi(g)=e\}$, and its image is $\phi(G)$. Both are subgroups by the subgroup test.

A homomorphism is injective exactly when its kernel is trivial: equality $\phi(a)=\phi(b)$ is equivalent to $\phi(a^{-1}b)=e$. An isomorphism is a bijective homomorphism; its inverse is also a homomorphism, so it identifies two group structures, not merely two sets of the same size.

Examples include $\mathbb Z\to\mathbb Z/n\mathbb Z$, $k\mapsto[k]$, and $\det:GL_n(\mathbb R)\to\mathbb R^\times$. The latter's kernel consists of determinant-one matrices. A map such as $x\mapsto x+1$ on the additive reals is bijective but not a homomorphism, because it does not preserve zero.

2 Normality makes multiplication well defined

A subgroup $N\le G$ is normal, written $N\trianglelefteq G$, when $gNg^{-1}=N$ for every $g\in G$. This is equivalent to $gN=Ng$. Every subgroup of an abelian group is normal. Every kernel is normal, since $\phi(gng^{-1})=\phi(g)e\phi(g)^{-1}=e$, and conjugating back supplies equality rather than mere inclusion.

Quotient theorem. If $N$ is normal, the rule $(aN)(bN)=abN$ gives a group structure on the cosets, denoted $G/N$.

Proof. If $a'=an_1$ and $b'=bn_2$, then $a'b'=ab(b^{-1}n_1b)n_2\in abN$ by normality. Thus the output coset is independent of representatives. Associativity follows from $G$; identity is $N$ and inverse of $aN$ is $a^{-1}N$. $\square$

Conversely, if the proposed multiplication is well defined, $nN=N$ implies $(nN)(gN)=gN$ for $n\in N$, so $ngN=gN$ and $g^{-1}ng\in N$. Applying the same inclusion with $g^{-1}$ gives normality. This proves the condition is necessary, not just a useful sufficient test.

3 The first isomorphism theorem

Theorem. A homomorphism $\phi:G\to K$ induces an isomorphism

$$G/\ker\phi\ \cong\ \phi(G),\qquad g\ker\phi\longmapsto\phi(g).$$

Proof. Two cosets agree exactly when $g^{-1}h\in\ker\phi$, equivalently $\phi(g)=\phi(h)$. This simultaneously proves well-definedness and injectivity. Every image element is hit, and $\phi(gh)=\phi(g)\phi(h)$ proves multiplication is preserved. $\square$

For $\phi:\mathbb Z\to\mathbb Z/6\mathbb Z$, $k\mapsto[2k]$, the kernel is $3\mathbb Z$ and the image is $\{[0],[2],[4]\}$. Thus $\mathbb Z/3\mathbb Z$ is isomorphic to this image, not to the whole six-element target.

The sign map has kernel $A_n$, the even permutations. For $n\ge2$, it is onto $\{\pm1\}$, so $S_n/A_n\cong\{\pm1\}$. In particular, normality of $A_n$ follows from being a kernel rather than a separate coset enumeration.

4 Two further isomorphism theorems

Second theorem. If $H\le G$ and $N\trianglelefteq G$, then $HN=\{hn:h\in H,n\in N\}$ is a subgroup and

$$H/(H\cap N)\cong HN/N.$$

To prove subgroup closure, move an $N$ factor past an $H$ factor using normality; inverses are handled the same way. Restrict the projection $G\to G/N$ to $H$. Its kernel is $H\cap N$ and its image is $HN/N$, so the first theorem proves the result and also shows $H\cap N$ is normal in $H$.

Third theorem. If $N\trianglelefteq G$, $K\trianglelefteq G$, and $N\subseteq K$, then $K/N\trianglelefteq G/N$ and

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

The map $gN\mapsto gK$ is well defined because $N\subseteq K$, is onto, and has kernel $K/N$. Apply the first theorem. The statement does not omit normality hypotheses: without them the displayed quotients may not be groups.

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.

5 Graded practice

E1 Foundation. Prove every subgroup of index 2 is normal.

E2 First theorem. For $\phi:\mathbb Z\to\mathbb Z/12\mathbb Z$, $k\mapsto[8k]$, find kernel, image and the induced isomorphism.

E3 Third theorem. Identify $(\mathbb Z/12\mathbb Z)/(4\mathbb Z/12\mathbb Z)$.

E4 Nonexample. Prove $H=\{e,(1\ 2)\}$ is not normal in $S_3$ by conjugation.

6 Full solutions

S1. For $g\in H$, both $gH$ and $Hg$ equal $H$. For $g\notin H$, the only other left coset and the only other right coset are both the set complement $G\setminus H$. Thus $gH=Hg$ for every $g$.

S2. $[8k]=[0]$ exactly when $12\mid8k$, equivalently $3\mid k$. The kernel is $3\mathbb Z$ and image is $\{[0],[8],[4]\}$. The induced isomorphism is $\mathbb Z/3\mathbb Z\to\phi(\mathbb Z)$, $[k]_3\mapsto[8k]_{12}$.

S3. Use $N=12\mathbb Z\subset K=4\mathbb Z\subset G=\mathbb Z$. The third theorem gives $\mathbb Z/4\mathbb Z$. The intermediate subgroup has three elements, leaving four quotient cosets, consistent with the result.

S4. Conjugating $(1\ 2)$ by $(1\ 3)$ produces $(3\ 2)$, not in $H$. Thus $gHg^{-1}\ne H$ for that $g$. Lagrange gives an index of 3, not the special index-2 condition.

Summary

A kernel identifies exactly the differences a homomorphism erases. Normality makes coset multiplication independent of representatives. The first isomorphism theorem identifies the remaining information with the image; the second and third follow by applying it to carefully chosen projection maps.