Normal Subgroups

Normal subgroups are the mechanism behind quotient groups, which recur throughout mathematics: quotienting $\mathbb{Z}$ by $n\mathbb{Z}$ produces the cyclic group $\mathbb{Z}/n\mathbb{Z}$ used everywhere in number theory and cryptography (e.g. RSA arithmetic modulo $n$, and Diffie–Hellman key exchange in $(\mathbb{Z}/p\mathbb{Z})^\times$). In Galois theory, normal subgroups of a Galois group correspond precisely to intermediate fields that are themselves normal (Galois) extensions of the base field, which is the origin of the terminology. In crystallography and physics, normal subgroups of space groups (such as translation subgroups) let physicists pass to point groups describing the symmetry of a crystal lattice up to translation, essential in solid-state physics and materials science. In coding theory, normal subgroups underlie the construction of quotient codes and coset codes used in error-correcting code design. In computer science and computational group theory, algorithms for testing normality and computing quotient groups (implemented in systems like GAP and Magma) are basic primitives for symbolic computation with finite groups, used in cryptanalysis and combinatorial search. More generally, whenever an equivalence relation on a group is required to be compatible with the group operation — the defining feature of a congruence — the equivalence classes of the identity form exactly a normal subgroup, making this notion the group-theoretic instance of the universal algebraic concept of a congruence, mirrored by ideals in ring theory and normal subgroups' analogues in other algebraic categories.

:::quiz[If $[G:H]=2$, which statement is guaranteed true?]

• $H$ is abelian
• $H$ is the trivial subgroup
• $H \trianglelefteq G$ ✓
• $G$ is abelian

Ask the tutor: How does the well-definedness of coset multiplication become the group operation on $G/H$ once $H$ is normal, and why is the quotient map $G \to G/H$ then a homomorphism with kernel exactly $H$?

Think of conjugation $x \mapsto gxg^{-1}$ as a symmetry of $G$: it relabels elements while preserving the multiplication table. A subgroup $H$ is normal precisely when this relabeling always sends $H$ back onto itself, never onto some genuinely different subset. Ordinary subgroups can be moved around by conjugation into "different-looking copies"; normal subgroups are the ones that are conjugation-invariant, sitting inside $G$ the way a sphere sits inside space — the same from every rotated point of view. The coset picture makes the payoff concrete. Left cosets $\{gH\}$ partition $G$, and so do right cosets $\{Hg\}$, but in general the two partitions differ. Normality says these two partitions coincide, coset by coset. Once that happens, you can unambiguously multiply "coset classes" by multiplying representatives, because it no longer matters whether you track left or right translation. This is the razor's edge between "just a subgroup" and "a subgroup fit to build a quotient group from."

Example 1: $A_4 \trianglelefteq S_4$, checked three independent ways, and a subgroup of $A_4$ that is not normal in $A_4$. Let $G = S_4$, the symmetric group on $\{1,2,3,4\}$, with $|G| = 24$, and let $H = A_4$, the alternating group, with $|A_4| = 12$, so $[G:H] = 2$. Via the index-2 criterion: by the theorem, any subgroup of index 2 is automatically normal. So $A_4 \trianglelefteq S_4$ immediately. Via conjugation directly: conjugation by any $g \in S_4$ preserves cycle type (a standard fact: if $\sigma$ has cycle type $(c_1, c_2, \dots)$ then $g\sigma g^{-1}$ has the identical cycle type, since conjugation just relabels the symbols acted upon). Even permutations are exactly the permutations whose cycle type contributes an even number of transpositions when factored — equivalently, sign is a class function under conjugation because $\mathrm{sgn}(g\sigma g^{-1}) = \mathrm{sgn}(g)\mathrm{sgn}(\sigma)\mathrm{sgn}(g)^{-1} = \mathrm{sgn}(\sigma)$ (using that $\mathrm{sgn}$ is a homomorphism into the abelian group $\{\pm1\}$, so its values commute). Hence conjugation sends even permutations to even permutations: $g A_4 g^{-1} \subseteq A_4$ for all $g$, and by the equivalence theorem this gives $A_4 \trianglelefteq S_4$. Via kernels: $A_4 = \ker(\mathrm{sgn})$ where $\mathrm{sgn}: S_4 \to \{\pm 1\}$ is the sign homomorphism. By the Kernels-are-Normal corollary, $A_4 \trianglelefteq S_4$ automatically. Now consider $K = \{e, (12)(34)\} \le A_4$, a subgroup of order 2. Is $K \trianglelefteq A_4$? Conjugate $(12)(34)$ by $g = (123) \in A_4$:

Compute this by tracking where each symbol goes (apply right-to-left: first $(132)$, then $(12)(34)$, then $(123)$). Trace $1$: $(132)$ sends $1\mapsto 3$; $(12)(34)$ sends $3 \mapsto 4$; $(123)$ sends $4 \mapsto 4$. So $1 \mapsto 4$. Trace $2$: $(132)$ sends $2 \mapsto 1$; $(12)(34)$ sends $1 \mapsto 2$; $(123)$ sends $2 \mapsto 3$. So $2 \mapsto 3$. Trace $3$: $(132)$ sends $3\mapsto 2$; $(12)(34)$ sends $2\mapsto 1$; $(123)$ sends $1 \mapsto 2$. So $3 \mapsto 2$. Trace $4$: $(132)$ sends $4 \mapsto 4$; $(12)(34)$ sends $4 \mapsto 3$; $(123)$ sends $3\mapsto 1$. So $4 \mapsto 1$. Reading off the images $1\to4,\,2\to3,\,3\to2,\,4\to1$, the resulting permutation is $(14)(23)$. Since $(14)(23) \notin K = \{e,(12)(34)\}$, we have found $g \in A_4$ with $gKg^{-1} \not\subseteq K$. Hence $K$ is not normal in $A_4$, even though $A_4$ itself is normal in the larger group $S_4$ — normality is not automatically inherited by subgroups of a normal subgroup, a genuinely important cautionary phenomenon.

Example 2: A concrete failure and success of coset multiplication in $S_3$. Let $G = S_3 = \{e, (12), (13), (23), (123), (132)\}$, $|G|=6$. Non-normal case. Let $H = \{e, (12)\}$, a subgroup of order 2, index 3. The left cosets are

(using the convention that permutations compose left-to-right acting on the right, i.e. $(13)(12)$ means apply $(12)$ first, then $(13)$: trace $1\to2\to2$, $2\to1\to3$, $3\to3\to1$, giving $(1 2 3)$ wait — let us fix composition convention explicitly as functions, applying the rightmost permutation first. Then $(13)(12)$: apply $(12)$ first: $1\to2, 2\to1, 3\to3$; then apply $(13)$: $2\to2,1\to3,3\to1$. Composite: $1\to2\to2$, $2\to1\to3$, $3\to3\to1$. So $(13)(12) = (1\,2\,3)$ hold on, tracking again carefully: composite sends $1\mapsto 2$, $2 \mapsto 3$, $3\mapsto 1$, which is the 3-cycle $(123)$.) So $(13)H = \{(13),(123)\}$. Now compute the right coset $H(13) = \{(13), (12)(13)\}$. Compute $(12)(13)$ (apply $(13)$ first, then $(12)$): $1\to3\to3$, $3\to1\to2$, $2\to2\to1$. This sends $1\to3, 3\to2, 2\to1$, i.e. the 3-cycle $(1\,3\,2) = (132)$. So $H(13) = \{(13),(132)\}$. Compare: $(13)H = \{(13),(123)\}$ while $H(13) = \{(13),(132)\}$. These are different sets since $(123) \ne (132)$. So the left coset $(13)H$ is not the right coset $H(13)$, confirming (via criterion (3)/(4) of the equivalence theorem) that $H$ is not normal in $S_3$ — consistent with $H$ being a subgroup of order 2 fixing only one of three points, which is manifestly moved by conjugation. Because $H$ is not normal, coset multiplication genuinely breaks: take the coset $(13)H = \{(13),(123)\}$ and multiply "by representative $(13)$" against itself: $(13)H \cdot (13)H \overset{?}{=} \big((13)(13)\big)H = eH = H$. But redo the computation with the alternate representative $(123) \in (13)H$: $(123)H\cdot(123)H = \big((123)(123)\big)H = (132)H$. Compute $(132)H = \{(132),(132)(12)\}$; find $(132)(12)$ (apply $(12)$ first, then $(132)$): $1\to2\to1$, $2\to1\to3$, $3\to3\to2$, giving the 3-cycle $(23)$... tracking again: $1\to1$, $2\to3$, $3\to2$, so this is the transposition $(23)$. So $(132)H = \{(132),(23)\}$. We now have two different answers for "$(13)H$ times itself": using representative $(13)$ we got $H = \{e,(12)\}$, and using representative $(123)$ we got $(132)H = \{(132),(23)\}$. These are different cosets, so the operation $(aH)(bH):=(ab)H$ genuinely gives different results depending on the representative — exactly the failure predicted by the theorem, since $H$ is not normal. Normal case, for contrast. Let $N = \{e,(123),(132)\} \le S_3$, index 2, hence normal by the index-2 criterion. Its two cosets are $N$ and $S_3\setminus N = \{(12),(13),(23)\}$. One checks directly that any representative of the non-identity coset, multiplied by any representative of either coset, lands in the coset predicted by the "sign" of the representatives (transpositions are odd, 3-cycles are even), so the multiplication is consistent — this is exactly the two-element quotient $G/N \cong \mathbb{Z}/2\mathbb{Z}$ that normality guarantees is well-defined.