A subgroup can fail to be normal in a spectacular way. Take the subgroup $H = \langle (1\ 2) \rangle$ of the symmetric group $S_3$, generated by the transposition that swaps $1$ and $2$. This is a perfectly legitimate subgroup of order $2$, but conjugating it by the permutation $\sigma = (1\ 2\ 3)$ gives \begin{align*} \sigma H \sigma^{-1} &= \{e,\ \sigma (1\ 2) \sigma^{-1}\} = \{e,\ (2\ 3)\}, \end{align*} which is a completely different subgroup of $S_3$. The group $S_3$ acts on its own subgroups by conjugation, and $H$ is not a fixed point of that action. This failure has consequences: you cannot form the quotient $S_3 / H$ in any sensible way, because the left and right cosets of $H$ do not agree, and so "cosets" do not multiply consistently. The concept of a normal subgroup exists precisely to identify which subgroups are immune to conjugation — the ones for which $gHg^{-1} = H$ for every $g \in G$ — and these are the subgroups that permit a well-defined quotient group. [example: Failure of Left-Right Coset Agreement in $S_3$] To see concretely why non-normality obstructs the quotient construction, work out the cosets of $H = \{e, (1\ 2)\}$ in $S_3$. The left cosets are \begin{align*} eH &= \{e,\ (1\ 2)\}, \\ (1\ 2\ 3)H &= \{(1\ 2\ 3),\ (1\ 3)\}, \\ (1\ 3\ 2)H &= \{(1\ 3\ 2),\ (2\ 3)\}. \end{align*} The right cosets are \begin{align*} He &= \{e,\ (1\ 2)\}, \\ H(1\ 2\ 3) &= \{(1\ 2\ 3),\ (2\ 3)\}, \\ H(1\ 3\ 2) &= \{(1\ 3\ 2),\ (1\ 3)\}. \end{align*} The left coset $(1\ 2\ 3)H = \{(1\ 2\ 3),\ (1\ 3)\}$ and the right coset $H(1\ 2\ 3) = \{(1\ 2\ 3),\ (2\ 3)\}$ are not equal: the element $(1\ 3)$ appears in the left coset but not in the right. When left and right cosets differ, there is no natural group operation on the set of cosets: the "product" of two cosets would depend on which coset representative you choose, not just on the coset itself. [/example] Normal subgroups are the correct level of generality for forming quotient groups, and they appear throughout algebra in roles that may not be immediately obvious — as kernels of homomorphisms, as building blocks in the composition series that classify groups up to extensions, and as the inputs to direct and semidirect product constructions. ## Definition Before stating the definition, it is worth understanding what condition we actually need. We want to form a quotient $G/N$ whose elements are cosets and whose multiplication is \begin{align*} (gN)(hN) &:= (gh)N. \end{align*} For this to be well-defined, if we replace $g$ by $gn_1$ and $h$ by $hn_2$ for any $n_1, n_2 \in N$, the product must still land in the same coset: \begin{align*} (gn_1)(hn_2) &= g(n_1 h)n_2 = g(h \cdot h^{-1} n_1 h) n_2 \in ghN \end{align*} provided $h^{-1} n_1 h \in N$ for every $h \in G$ and every $n_1 \in N$. The condition $h^{-1}Nh \subset N$ for all $h \in G$ is exactly what we need — and combined with the reverse inclusion obtained by applying the condition to $h^{-1}$, this gives $h^{-1}Nh = N$. That is the definition. [definition: Normal Subgroup] Let $G$ be a group and $N \le G$ a subgroup. We say $N$ is a **normal subgroup** of $G$, written $N \trianglelefteq G$, if \begin{align*} gNg^{-1} &= N \quad \text{for all } g \in G, \end{align*} where $gNg^{-1} := \{gng^{-1} : n \in N\}$ denotes the conjugate of $N$ by $g$. If additionally $N \ne G$, we say $N$ is a **proper normal subgroup**, written $N \triangleleft G$. [/definition] Several equivalent characterizations of normality are worth recording immediately, because each reveals a different face of the condition. [quotetheorem:787] The equivalence of (i) and (ii) is slightly subtle: if $gNg^{-1} \subset N$ for all $g$, applying this to $g^{-1}$ gives $g^{-1}Ng \subset N$, i.e., $N \subset gNg^{-1}$, which together with the original inclusion forces equality. Condition (iii) is what makes the quotient construction work. Condition (v) gives an intrinsic, coordinate-free description: $N$ is normal in $G$ if and only if knowing that $n \in N$ forces the entire conjugacy class $\operatorname{Cl}(n) = \{gng^{-1} : g \in G\}$ to lie inside $N$. [remark: Notation] We write $N \trianglelefteq G$ for normal subgroups and $N \triangleleft G$ for proper normal subgroups. The symbol $\le$ is reserved for arbitrary subgroups. An abelian group is its own center, so every subgroup of an abelian group is normal — condition (iii) is immediate, since $gN = Ng$ holds elementwise when $G$ is abelian. [/remark] ## The Quotient Group Construction With the definition in hand, we can verify that normality is exactly the hypothesis needed to make the set of cosets into a group. To appreciate why this theorem is nontrivial, consider what could go wrong. The set $G/N = \{gN : g \in G\}$ of left cosets is well-defined regardless of whether $N$ is normal. We want to define a binary operation by $(gN)(hN) := (gh)N$. The danger is that a coset has many representatives: $gN = g'N$ whenever $g' = gn$ for some $n \in N$. The product must not depend on which representatives we pick. [quotetheorem:790] The content of the theorem is the well-definedness: if $gN = g'N$ and $hN = h'N$, then $g' = gn_1$ and $h' = hn_2$ for some $n_1, n_2 \in N$, and \begin{align*} g'h' &= (gn_1)(hn_2) = g(n_1 h)n_2 = gh(h^{-1}n_1 h)n_2. \end{align*} Since $N \trianglelefteq G$, we have $h^{-1}n_1 h \in N$, so $(h^{-1}n_1 h)n_2 \in N$, which gives $g'h'N = ghN$. Normality is not just sufficient here — it is necessary. Suppose coset multiplication is well-defined, and pick any $n \in N$ and $g \in G$. Since $n \in N$, the coset $nN$ equals $N = eN$, so the product $(nN)(gN)$ must equal $(eN)(gN) = gN$ no matter which representative of $nN$ we pick. Choosing $n$ itself as the representative gives $(ng)N = gN$, hence $g^{-1}(ng) = g^{-1}ng \in N$. Since $n$ and $g$ were arbitrary, this is the normality condition. [example: The Quotient $\mathbb{Z}/n\mathbb{Z}$] The integers $(\mathbb{Z}, +)$ form an abelian group. The subgroup $n\mathbb{Z} = \{nk : k \in \mathbb{Z}\}$ of multiples of $n$ is normal (as every subgroup of an abelian group is). The coset of $a \in \mathbb{Z}$ is \begin{align*} a + n\mathbb{Z} &= \{a + nk : k \in \mathbb{Z}\}, \end{align*} which is the residue class of $a$ modulo $n$. The quotient group $\mathbb{Z}/n\mathbb{Z}$ has $n$ elements: the cosets $0 + n\mathbb{Z},\ 1 + n\mathbb{Z},\ \ldots,\ (n-1) + n\mathbb{Z}$. The group operation is \begin{align*} (a + n\mathbb{Z}) + (b + n\mathbb{Z}) &= (a + b) + n\mathbb{Z}, \end{align*} which is precisely addition modulo $n$. This is the cyclic group $C_n$ in its most concrete incarnation. [/example] [illustration:cosets-of-nz-collapse] [example: The Alternating Group $A_n$ is Normal in $S_n$] The alternating group $A_n \le S_n$ consists of all even permutations. It has index $2$ in $S_n$, since every permutation is either even or odd, and the even permutations form a subgroup while the odd ones do not. Any subgroup of index $2$ is normal: if $[G:H] = 2$, there are exactly two left cosets ($H$ and $G \setminus H$) and exactly two right cosets ($H$ and $G \setminus H$), so $gH = G \setminus H = Hg$ for any $g \notin H$. Applied to $A_n \le S_n$, we get $A_n \trianglelefteq S_n$, and the quotient \begin{align*} S_n / A_n &\cong \mathbb{Z}/2\mathbb{Z} \end{align*} is the group with two elements, where a permutation maps to $0$ if it is even and to $1$ if it is odd. This quotient is the sign homomorphism $\operatorname{sgn}: S_n \to \{-1, +1\}$ made into an isomorphism of quotients. [/example] ## Normality and Homomorphisms The deepest source of normal subgroups in practice is not geometric intuition about conjugation, but algebra: kernels of group homomorphisms are always normal, and conversely every normal subgroup is a kernel. [definition: Kernel of a Homomorphism] Let $\varphi: G \to H$ be a group homomorphism. The **kernel** of $\varphi$ is the subgroup \begin{align*} \ker \varphi &:= \{g \in G : \varphi(g) = e_H\}, \end{align*} where $e_H$ is the identity element of $H$. [/definition] Having defined the kernel, the natural question is whether it is always a normal subgroup — and the answer is yes, by a short computation that uses only the homomorphism property. [quotetheorem:788] The proof is a direct computation: if $n \in \ker \varphi$ and $g \in G$, then $\varphi(gng^{-1}) = \varphi(g)\varphi(n)\varphi(g^{-1}) = \varphi(g) \cdot e_H \cdot \varphi(g)^{-1} = e_H$, so $gng^{-1} \in \ker \varphi$. The converse — that every normal subgroup is a kernel — is the content of the First Isomorphism Theorem, and it makes the notion of normality completely equivalent to the notion of "kernel of some homomorphism." [quotetheorem:842] This theorem is the fundamental tool for identifying quotient groups. Every time you write down a surjective homomorphism $\varphi: G \to Q$, you have realized $Q$ as the quotient $G / \ker \varphi$. In the example $\operatorname{sgn}: S_n \to \{-1, 1\}$, the kernel is $A_n$ and the image is $\{-1, 1\} \cong \mathbb{Z}/2\mathbb{Z}$, recovering $S_n / A_n \cong \mathbb{Z}/2\mathbb{Z}$. [example: The Center as a Normal Subgroup via an Action] The center $Z(G) = \{g \in G : gx = xg \text{ for all } x \in G\}$ is always normal in $G$. One way to see this: define the conjugation action of $G$ on itself by \begin{align*} G \times G &\to G \\ (g, x) &\mapsto gxg^{-1}. \end{align*} This gives a group homomorphism $\varphi: G \to \operatorname{Aut}(G)$ by $\varphi(g)(x) = gxg^{-1}$, called the conjugation homomorphism or the map to inner automorphisms. An element $g \in G$ lies in $\ker \varphi$ if and only if $gxg^{-1} = x$ for all $x \in G$, i.e., if and only if $g \in Z(G)$. Since kernels of homomorphisms are normal, $Z(G) = \ker \varphi \trianglelefteq G$. Moreover, the image of $\varphi$ is the group of inner automorphisms $\operatorname{Inn}(G) \le \operatorname{Aut}(G)$, and the First Isomorphism Theorem gives $G / Z(G) \cong \operatorname{Inn}(G)$. [/example] ## Simple Groups and Composition Series Once we understand normal subgroups and quotients, a natural question arises: can every group be built from simpler pieces? The right framework is the composition series, which requires understanding which groups cannot be simplified further. A group with no proper normal subgroups (other than the trivial group $\{e\}$ and $G$ itself) cannot be further decomposed by the quotient construction. These are the irreducible building blocks of group theory. [definition: Simple Group] A group $G$ is **simple** if $|G| > 1$ and the only normal subgroups of $G$ are $\{e\}$ and $G$ itself. [/definition] The condition $|G| > 1$ excludes the trivial group, which would otherwise be a degenerate simple group. Every group of prime order $p$ is simple: by Lagrange's theorem, the only subgroups have orders $1$ and $p$, and with a group of prime order being cyclic, these are also the only normal subgroups. The alternating group $A_n$ for $n \ge 5$ is simple — this is a classical theorem, and its proof is one of the central achievements of nineteenth-century algebra. The simplicity of $A_5$ is why the general quintic polynomial has no solution by radicals: Galois theory translates solvability by radicals into the existence of a composition series with abelian factors, and the simple non-abelian group $A_5$ cannot appear in any such series. The cutoff $n \ge 5$ is structurally meaningful. For $n = 3$, the group $A_3 \cong \mathbb{Z}/3\mathbb{Z}$ is simple, but only as a cyclic group of prime order — abelian simplicity, not the deeper non-abelian kind. For $n = 4$, the group $A_4$ is not simple: the Klein four-group $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ is a proper normal subgroup of $A_4$. Simple non-abelian alternating groups first appear at $n = 5$. [quotetheorem:849] The key steps of the proof are: 3-cycles generate $A_n$ for $n \ge 3$; any nontrivial normal subgroup $N \trianglelefteq A_n$ for $n \ge 5$ can be shown to contain a 3-cycle (by exploiting the freedom afforded by $n \ge 5$); and since $N$ is then closed under conjugation by all of $A_n$ and 3-cycles generate $A_n$, this forces $N = A_n$. A complete proof is given on the [Simplicity of the Alternating Group](/page/Simplicity%20of%20the%20Alternating%20Group) page. With simple groups identified as the atoms, the Jordan–Hölder theorem tells us that every finite group has a canonical decomposition into simple groups, up to ordering and isomorphism. [definition: Composition Series] A **composition series** of a group $G$ is a finite chain of subgroups \begin{align*} \{e\} = G_0 \trianglelefteq G_1 \trianglelefteq G_2 \trianglelefteq \cdots \trianglelefteq G_k = G \end{align*} such that each quotient $G_{i+1}/G_i$ is simple. The quotients $G_{i+1}/G_i$ are called the **composition factors** of the series. [/definition] Different choices of intermediate subgroups can produce different composition series for the same group. The fundamental question is whether the simple pieces that appear — the composition factors — are canonical, or whether they depend on the choices made. The Jordan–Hölder theorem answers this: the composition factors are an invariant of the group. [quotetheorem:2625] The Jordan–Hölder theorem reduces the classification of finite groups to two sub-problems: classify all finite simple groups (a program completed in 2004, running to tens of thousands of pages), and understand all possible extensions — that is, all ways to build a group $G$ with a normal subgroup $N$ such that $N \cong A$ and $G/N \cong B$ for given simple groups $A$ and $B$. The extension problem is generally unsolved. ## Normality and Group Actions Normal subgroups interact naturally with group actions. Suppose $G$ acts on a set $X$ and $N \trianglelefteq G$. Then $N$ also acts on $X$, and the orbits of $N$ are permuted by $G$ (since $N$ is normal, the set of $N$-orbits is preserved by the $G$-action). This allows the quotient $G/N$ to act on the set of $N$-orbits, which is a fundamental technique in representation theory and the study of group extensions. A particularly concrete instance is the action of a group on itself by conjugation. [definition: Conjugacy Class] Let $G$ be a group and $g \in G$. The **conjugacy class** of $g$ is \begin{align*} \operatorname{Cl}(g) &:= \{hgh^{-1} : h \in G\}. \end{align*} [/definition] The conjugacy classes partition $G$ (they are the orbits of the conjugation action $G \curvearrowright G$ defined by $(h, g) \mapsto hgh^{-1}$). A subgroup $N \le G$ is normal if and only if it is a union of conjugacy classes — a rephrasing of condition (v) from the Equivalent Conditions theorem above. This perspective is especially useful for finite groups, because the conjugacy classes can be computed explicitly, and any union of conjugacy classes containing the identity is a candidate for a normal subgroup. [example: Normal Subgroups of $S_4$ via Conjugacy Classes] The symmetric group $S_4$ has $24$ elements, partitioned into conjugacy classes by cycle type: \begin{align*} \operatorname{Cl}(e) &= \{e\}, & &|\operatorname{Cl}(e)| = 1, \\ \operatorname{Cl}((12)) &= \{(12),(13),(14),(23),(24),(34)\}, & &|\operatorname{Cl}((12))| = 6, \\ \operatorname{Cl}((123)) &= \{(123),(132),(124),(142),(134),(143),(234),(243)\}, & &|\operatorname{Cl}((123))| = 8, \\ \operatorname{Cl}((1234)) &= \{(1234),(1243),(1324),(1342),(1423),(1432)\}, & &|\operatorname{Cl}((1234))| = 6, \\ \operatorname{Cl}((12)(34)) &= \{(12)(34),(13)(24),(14)(23)\}, & &|\operatorname{Cl}((12)(34))| = 3. \end{align*} A normal subgroup of $S_4$ must be a union of conjugacy classes containing $e$. The sizes of unions containing $\{e\}$ that are also divisors of $24$ (since by Lagrange's theorem, $|N|$ divides $|G| = 24$) are: $1$, $1+3 = 4$, $1+3+8 = 12$, and $1+6+6+8+3 = 24$. The corresponding normal subgroups are: - $\{e\}$ (the identity subgroup), - $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ (the Klein four-group), - $A_4$ (the alternating group, consisting of even permutations — classes of $e$, $(123)$-type, and $(12)(34)$-type), - $S_4$ itself. We can verify $V_4 \trianglelefteq S_4$ independently: $V_4$ has index $6$ in $S_4$, and conjugation by any element of $S_4$ sends each non-identity element of $V_4$ to another non-identity element of $V_4$. As a representative computation, we conjugate $(12)(34)$ by $(123)$: \begin{align*} (123)(12)(34)(123)^{-1} &= (123)(12)(34)(132). \end{align*} We compute the composition right-to-left, tracking each of $1, 2, 3, 4$: - $1 \xrightarrow{(132)} 3 \xrightarrow{(12)(34)} 4 \xrightarrow{(123)} 4$. - $2 \xrightarrow{(132)} 1 \xrightarrow{(12)(34)} 2 \xrightarrow{(123)} 3$. - $3 \xrightarrow{(132)} 2 \xrightarrow{(12)(34)} 1 \xrightarrow{(123)} 2$. - $4 \xrightarrow{(132)} 4 \xrightarrow{(12)(34)} 3 \xrightarrow{(123)} 1$. The result sends $1 \mapsto 4, 2 \mapsto 3, 3 \mapsto 2, 4 \mapsto 1$, which is the permutation $(14)(23)$. This lies in $V_4$, confirming closure under this conjugation. The verifications for the other generators of $S_4$ — and for the other elements of $V_4$ — proceed analogously, since the cycle type $(2,2)$ is preserved by conjugation in $S_n$. The subgroup $V_4 \trianglelefteq A_4 \trianglelefteq S_4$ is the beginning of the composition series \begin{align*} \{e\} \trianglelefteq \mathbb{Z}/2\mathbb{Z} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4, \end{align*} with composition factors $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/3\mathbb{Z}$, and $\mathbb{Z}/2\mathbb{Z}$ — all abelian, which is why $S_4$ (unlike $S_5$) is solvable. [/example] [illustration:s4-normal-subgroup-chain] ## The Normalizer and When Subgroups Are Normal Not every subgroup is normal, and it is useful to have a measure of "how normal" a subgroup is — the largest subgroup of $G$ in which a given $H$ is normal. [definition: Normalizer] Let $G$ be a group and $H \le G$. The **normalizer** of $H$ in $G$ is \begin{align*} N_G(H) &:= \{g \in G : gHg^{-1} = H\}. \end{align*} [/definition] The normalizer $N_G(H)$ is always a subgroup of $G$ containing $H$. It is the largest subgroup of $G$ in which $H$ sits as a normal subgroup: $H \trianglelefteq N_G(H)$, and if $K \le G$ with $H \trianglelefteq K$, then $K \subset N_G(H)$. A subgroup $H$ is normal in $G$ if and only if $N_G(H) = G$. [explanation: The Normalizer as a Measure of Normality] The normalizer interpolates between $H$ and $G$. At one extreme, if $H \trianglelefteq G$, then $N_G(H) = G$: every element of $G$ normalizes $H$. At the other extreme, if $H$ is "as far from normal as possible," then $N_G(H) = H$: only the elements of $H$ itself preserve $H$ under conjugation. This happens, for instance, for a Sylow $p$-subgroup $P$ of a group where $P$ is self-normalizing: $N_G(P) = P$. The index $[G : N_G(H)]$ counts the number of distinct conjugates of $H$ in $G$ — that is, the size of the orbit $\{gHg^{-1} : g \in G\}$ under the conjugation action. If $N_G(H) = G$, there is only one conjugate (namely $H$ itself), confirming normality. If $[G : N_G(H)] = k$, then $H$ has exactly $k$ conjugates in $G$. This count is useful in proving that certain groups must have a normal subgroup when the number of conjugates of a Sylow subgroup is forced to be $1$ by arithmetic constraints — a central technique in the Sylow theorems. [/explanation] The following theorem makes the orbit count precise, giving a formula that directly connects the normalizer to the number of conjugates. [quotetheorem:2626] To see this theorem in action, we work out a case where the conjugate count forces a normal subgroup to exist. [example: Sylow Subgroups and Normality in a Group of Order 12] Let $G$ be a group of order $12 = 4 \cdot 3$. By Sylow's theorem, the number of Sylow $3$-subgroups (subgroups of order $3$) divides $4$ and is congruent to $1 \pmod 3$, so it is $1$ or $4$. If there is only one Sylow $3$-subgroup $P$, then $P$ is the unique subgroup of order $3$ and hence $gPg^{-1} = P$ for all $g \in G$ (since conjugation by $g$ sends a subgroup of order $3$ to a subgroup of order $3$, and there is only one). Thus $P \trianglelefteq G$. If there are four Sylow $3$-subgroups, each of order $3$ and cyclic, any two of them intersect in only the identity: $|P \cap Q|$ divides $|P| = 3$, so it is $1$ or $3$, and if it were $3$ then $P = Q$. Each contributes $2$ non-identity elements, giving $4 \times 2 = 8$ elements of order $3$ in $G$. This leaves $12 - 8 = 4$ elements not of order $3$, which must form the unique Sylow $2$-subgroup (order $4$). So the Sylow $2$-subgroup is normal in this case. Either way, $G$ has a normal Sylow subgroup — this reasoning shows that no group of order $12$ is simple. [/example] ## References Dummit, D. S. and Foote, R. M., *Abstract Algebra*, 3rd ed. (2004). Lang, S., *Algebra*, revised 3rd ed. (2002). Rotman, J. J., *An Introduction to the Theory of Groups*, 4th ed. (1995). Hungerford, T. W., *Algebra* (1974).