Conjugacy Classes and Normal Subgroups

Conjugation, the map $g\mapsto aga^{-1}$ , is a group acting on itself by inner automorphisms, and it answers a physicist's question: which group elements are "the same operation seen from a different frame"? A rotation by a quarter turn about the vertical axis and a quarter turn about a horizontal axis are not equal, yet they are manifestly the same kind of motion — one is the other viewed after a change of orientation. Conjugation makes that precise, and its orbits, the conjugacy classes, carve the group into the pieces that representation theory treats as indivisible. Characters are constant on these classes, the number of classes equals the number of irreducible representations, and so identifying the classes of a symmetry group is the concrete first move in building any character table. The dual object is the normal subgroup: a subgroup $N$ that conjugation carries onto itself, $gNg^{-1}=N$ for every $g$ . Equivalently — and this is the bridge — a subgroup is normal exactly when it is a union of whole conjugacy classes. That single condition is what lets the cosets of $N$ multiply consistently, manufacturing the quotient group $G/N$ in which $N$ has been collapsed to a point. Galois introduced these "invariant" subgroups to decide when a polynomial is solvable by radicals; physics inherited them as the machinery that splits a spectrum by parity, factors a symmetry group into simpler quotients, and organizes identical-particle states. The class equation $|G|=|Z(G)|+\sum_i[G:Z_G(g_i)]$ turns all of this into arithmetic — a counting identity strong enough to prove that every group of prime-power order has a nontrivial center.

Key takeaways

Two elements are conjugate, $h\sim g$ , if $h=aga^{-1}$ for some $a\in G$ ; conjugacy is an equivalence relation that partitions $G$ into disjoint conjugacy classes.
A class size equals the index of a centralizer, $|C(g)|=[G:Z_G(g)]=|G|/|Z_G(g)|$ , so every class size divides $|G|$ — the content of the class equation.
A subgroup is normal, $N\trianglelefteq G$ , precisely when $gNg^{-1}=N$ for all $g$ , equivalently when $N$ is a union of complete conjugacy classes.
Normality is exactly the condition that the cosets form a group: the quotient $G/N$ has a well-defined product $(g_1N)(g_2N)=(g_1g_2)N$ .
Kernels of homomorphisms are normal, and every normal subgroup is the kernel of its own canonical projection $G\to G/N$ .
In physics the classes label the columns of character tables, because characters are class functions: $\chi(aga^{-1})=\chi(g)$ .
Key idea: conjugacy classes partition a group into orbits under inner automorphism; a subgroup is normal exactly when it is a union of such classes, and that union-of-classes condition is precisely what makes the coset space a quotient group.

Where this lives today

The class structure of a symmetry group is the first thing a working spectroscopist computes. Every molecular point group — $C_{3v}$ for ammonia, $T_d$ for methane, $O_h$ for an octahedral complex — is tabulated by its conjugacy classes, because the columns of the character table are the classes and the rows are the irreducible representations, equal in number. Vibrational modes, molecular orbitals, and crystal-field splittings are then read off by decomposing a reducible representation into irreducibles using characters that, by construction, take one value per class. In solid-state physics the same machinery classifies the 230 crystallographic space groups: a space group has the pure translations as a normal subgroup, and the quotient by it is the point group — the normal-subgroup-and-quotient construction of this article is literally how the classification is organized. In particle physics, conjugation in $SU(2)$ and $SO(3)$ collapses a rotation's class to depend only on its angle $\theta$ , which is why the spin- $j$ character $\chi_j(\theta)=\sin\!\big((2j+1)\theta/2\big)/\sin(\theta/2)$ depends on that single variable, and why grand-unified model builders hunt for symmetry-breaking chains $G\to H$ in which $H$ is a subgroup selected by exactly which representations stay invariant. And the deepest modern echo is the classification of finite simple groups — the groups with no nontrivial normal subgroup — completed across thousands of journal pages and crowned by the Monster, whose 194 conjugacy classes turned out, in "monstrous moonshine", to encode the coefficients of a modular function from string theory.

The physical picture

Picture the six symmetries of an equilateral triangle. A reflection that fixes vertex 1 and a reflection that fixes vertex 2 are different operations — they move different points — but they are the same type of symmetry: each is a mirror flip through a vertex. Conjugation is the algebra of that intuition. If $a$ is the rotation that carries vertex 1 to vertex 2, then $a$ takes the whole "reflect through vertex 1" operation, performed in the original frame, into "reflect through vertex 2", performed in the rotated frame: $a\,\sigma_1\,a^{-1}=\sigma_2$ . The two reflections therefore sit in one conjugacy class. The first figure is exactly this — conjugation as relabelling the frame, turning one reflection into another of the same kind.

Figure

Conjugation is one operation seen in a relabelled frame: rotating the triangle by a carries the mirror σ₁ through vertex 1 into the mirror σ₂ through vertex 2, so a σ₁ a⁻¹ = σ₂ and both reflections lie in one conjugacy class

Once the group is sliced into classes, two facts drop out immediately. The center $Z(G)$ is the collection of elements that commute with everything, and each of them is its own class of size one — conjugation cannot move them. Everything else lives in larger classes whose sizes must divide $|G|$ . A subgroup is normal exactly when it is assembled from complete classes and never cuts one in half; that is the algebraic reason its left and right cosets coincide. The second figure makes this concrete on the smallest non-abelian group, $S_3$ , where the class equation reads $6=1+2+3$ and the alternating subgroup $A_3$ is a clean union of two classes.

Figure

The class equation of S₃ : the six elements split into three conjugacy classes of sizes 1, 2, 3 so that 6 = 1 + 2 + 3; the union of the size-1 and size-2 classes is A₃, a normal subgroup of index 2, while the two-element set e and (12) splits the size-3 class and is not normal

The core equations

The whole subject starts from a single relation. Two elements are conjugate when one is obtained from the other by an inner automorphism — a change of frame. This relation is reflexive, symmetric, and transitive, so it partitions the group.

h\sim g \iff \exists\,a\in G:\ h=a\,g\,a^{-1} \qquad (1)

where: $g,h$ — elements of the group $G$ ; $a$ — the conjugating element; $\sim$ — "is conjugate to", an equivalence relation. Reading it: $h$ is $g$ performed in the coordinate system that $a$ sets up, so conjugate elements are the same operation up to a relabelling. The orbit of $g$ under conjugation is its class, and the elements that leave $g$ fixed under conjugation form its centralizer, a genuine subgroup. These two sets are the raw material for counting.

C(g)=\{\,a\,g\,a^{-1}:a\in G\,\},\qquad Z_G(g)=\{\,a\in G:a\,g\,a^{-1}=g\,\} \qquad (2)

where: $C(g)$ — the conjugacy class (orbit) containing $g$ ; $Z_G(g)$ — the centralizer, the set of elements commuting with $g$ . Reading it: $g$ is central, $C(g)=\{g\}$ , exactly when $Z_G(g)=G$ . Applying the orbit–stabilizer theorem to the conjugation action ties the class size to the centralizer, and through Lagrange's theorem forces the class size to divide the group order.

|C(g)|=[G:Z_G(g)]=\frac{|G|}{|Z_G(g)|} \qquad (3)

where: $|C(g)|$ — number of elements in the class of $g$ ; $[G:Z_G(g)]$ — index of the centralizer. Reading it: a big class means a small centralizer; a class of size one means $g$ commutes with the whole group. Summing (3) over one representative from each class, and pulling the singleton central classes out of the sum, gives the class equation — the arithmetic backbone of finite group theory.

|G|=|Z(G)|+\sum_i[G:Z_G(g_i)] \qquad (4)

where: $Z(G)$ — the center, the union of all size-one classes; $g_i$ — one representative of each non-central class. Reading it: the group order is the number of central elements plus a sum of divisors of $|G|$ , each greater than one. Normality is the invariance of a subgroup under every inner automorphism, and it is equivalent to the geometric statement that $N$ is a union of complete classes.

N\trianglelefteq G \iff g\,N\,g^{-1}=N\ \ \forall g\in G \iff N=\bigsqcup_i C(g_i) \qquad (5)

where: $N$ — a subgroup of $G$ ; $\trianglelefteq$ — "is a normal subgroup of"; $\bigsqcup$ — disjoint union of whole conjugacy classes. Reading it: to test normality, check whether $N$ ever contains part of a class without containing all of it. When $N$ is normal, its cosets multiply by a representative-independent rule, and the resulting quotient is a group in its own right.

(g_1 N)(g_2 N)=(g_1 g_2)\,N \qquad (6)

where: $g_iN$ — a left coset of $N$ , one element of $G/N$ ; $G/N$ — the quotient group of cosets. Reading it: collapse $N$ to the identity and the leftover structure still multiplies coherently. Finally, every homomorphism hands you a normal subgroup for free — its kernel — and conversely every normal subgroup arises this way from its own projection.

\ker\varphi=\{\,g\in G:\varphi(g)=e'\,\}\ \trianglelefteq\ G \qquad (7)

where: $\varphi$ — a homomorphism $G\to G'$ ; $e'$ — the identity of the target group. Reading it: kernels and normal subgroups are the same objects viewed from two directions, the first isomorphism theorem being the bridge $G/\ker\varphi\cong\operatorname{im}\varphi$ .

Derivation AClass size equals the index of the centralizer

The goal is to prove the orbit–stabilizer identity $|C(g)|=[G:Z_G(g)]$ directly for the conjugation action, and to draw its immediate corollary that every conjugacy-class size divides the group order. This single counting fact is what turns the abstract partition-into-classes into the concrete, testable class equation, and it is the reason class sizes of a group of order eight can only be one, two, or four and never anything else.

Step 1. Fix $g$ and consider the map from group elements to conjugates, $a\mapsto aga^{-1}$ . Two elements $a$ and $b$ produce the same conjugate precisely when $aga^{-1}=bgb^{-1}$ , that is when $b^{-1}a$ commutes with $g$ , i.e. $b^{-1}a\in Z_G(g)$ , which says $a$ and $b$ lie in the same left coset of the centralizer. Why: it converts "how many distinct conjugates" into "how many cosets of $Z_G(g)$ ", replacing a hard count with a coset count.

Step 2. Therefore the assignment $aZ_G(g)\mapsto aga^{-1}$ is a well-defined bijection from the left cosets of $Z_G(g)$ onto the class $C(g)$ . It is onto by definition of the class and one-to-one by Step 1, so the number of conjugates equals the number of cosets: $|C(g)|=[G:Z_G(g)]$ . Why: a bijection is the cleanest possible proof of an equality of sizes, and it exposes the class as a coset space in disguise.

Step 3. By Lagrange's theorem $[G:Z_G(g)]=|G|/|Z_G(g)|$ is an integer dividing $|G|$ , so every class size divides the group order. Since $g\in Z_G(g)$ always, the centralizer is never empty, so no class can have size $|G|$ unless $Z_G(g)$ is trivial, which is impossible for $|G|>1$ . Why: this divisibility is the constraint that makes the class equation a rigid arithmetic identity rather than a loose sum. *As a quick check: for $S_3$ the formula gives class sizes $6/6=1$ , $6/2=3$ , and $6/3=2$ , matching the figure.*

Go deeper · Derivation B · 1 minCoset multiplication is well-defined if and only if N is normal

Derivation BCoset multiplication is well-defined if and only if N is normal

The aim is to show that the product of two cosets, defined by multiplying representatives, is independent of which representatives you pick exactly when $N$ is invariant under conjugation — and that this well-definedness is precisely what promotes the coset space $G/N$ into a group. This is the theorem that makes "quotienting out a normal subgroup" a legal operation, and it is why the kernel of every homomorphism must be normal.

Step 1. Define the product of cosets set-theoretically as $(g_1N)(g_2N)=\{\,g_1 n_1 g_2 n_2:n_1,n_2\in N\,\}$ , and demand that this whole set collapse to the single coset $(g_1 g_2)N$ . Why: for $G/N$ to be a group the product of two elements must be one element, i.e. one coset, not a scattered subset.

Step 2. Insert $g_2^{-1}g_2=e$ in the middle: $g_1 n_1 g_2 n_2=g_1 g_2\,(g_2^{-1} n_1 g_2)\,n_2$ . The product lands in $(g_1 g_2)N$ if and only if the conjugated element $g_2^{-1} n_1 g_2$ lies in $N$ for all $n_1\in N$ . Why: it isolates the one obstruction — a conjugate of an element of $N$ that might escape $N$ — and shows the whole question hinges on it.

Step 3. Requiring $g^{-1}Ng\subseteq N$ for every $g$ (and, applying it to $g^{-1}$ , the reverse inclusion too) is exactly $gNg^{-1}=N$ , the definition of normality. Conversely, if the product is well-defined for all representatives, choosing $g_1=e$ forces $n_1 g_2\in g_2 N$ , i.e. $g_2^{-1}n_1 g_2\in N$ — normality again. Why: it closes both directions, so well-definedness and normality are logically identical, not merely compatible.

Step 4. With normality in hand, associativity, the identity $eN=N$ , and inverses $(gN)^{-1}=g^{-1}N$ all descend directly from the corresponding facts in $G$ , so $G/N$ is a group. Why: the group axioms for the quotient are inherited for free once the product is legal, which is the entire payoff. *As a concrete illustration: for $N=A_3\trianglelefteq S_3$ the two cosets $A_3$ and $(12)A_3$ multiply as the group of order two, $S_3/A_3\cong\mathbb{Z}_2$ .*

Go deeper · Derivation C · 1 minThe class equation forces a nontrivial center for p-groups

Derivation CThe class equation forces a nontrivial center for p-groups

The goal is to prove that any group whose order is a prime power $p^n$ with $n\ge 1$ has a center strictly larger than the identity, using nothing but the class equation and the divisibility of class sizes. This theorem is the seed of the entire structure theory of $p$ -groups — it is what makes them solvable, nilpotent, and always built up from central extensions — and it is a clean showcase of the class equation doing real work.

Step 1. Write the class equation for $|G|=p^n$ : $p^n=|Z(G)|+\sum_i[G:Z_G(g_i)]$ , where the sum runs only over representatives of the non-central classes. Why: separating the singleton central classes from the rest is what lets the two sides be compared modulo $p$ .

Step 2. For each non-central $g_i$ the centralizer $Z_G(g_i)$ is a proper subgroup, so its index $[G:Z_G(g_i)]$ divides $p^n$ and is greater than one — hence a positive power of $p$ , and in particular divisible by $p$ . Why: every non-central term is a multiple of $p$ , so the entire sum vanishes modulo $p$ .

Step 3. Reduce the class equation modulo $p$ : the left side $p^n\equiv 0$ and the non-central sum $\equiv 0$ , forcing $|Z(G)|\equiv 0\pmod p$ . Since the identity is always central, $|Z(G)|\ge 1$ ; combined with $p\mid|Z(G)|$ this gives $|Z(G)|\ge p>1$ . Why: the arithmetic alone rules out a trivial center, with no need to know the group's multiplication table. *The worked example below confirms this for $D_4$ , where $|Z(D_4)|=2$ .*

Limiting cases

Conjugacy and normality either collapse to triviality or sharpen to a rigid constraint in the following special regimes.

Case	Result	Meaning
Abelian group	$aga^{-1}=g\ \ \forall a,g$	Every class is a singleton; every subgroup is normal; the class equation degenerates to $\vert G\vert=\sum 1$ .
Central element	$z\in Z(G)\iff C(z)=\{z\}$	Central elements are exactly the size-one classes, and $Z(G)$ is always a normal subgroup.
Index-2 subgroup	$[G:H]=2\implies H\trianglelefteq G$	A subgroup with only two cosets is automatically normal — the non-identity left and right cosets are both the complement.
Prime-power order	$\|G\|=p^n\implies\|Z(G)\|>1$	The class equation forbids a trivial center; $p$ -groups always have central elements to quotient by.
Simple group	only $\{e\}$ and $G$ are normal	No nontrivial proper quotient exists; simple groups are the indivisible building blocks of every finite group.

Common trapNormality is a relation, not a property of the subgroup alone

Two traps hide in the word "normal". First, normality is not transitive: $N$ can be normal in $H$ and $H$ normal in $G$ without $N$ being normal in $G$ — for example in $D_4$ , a reflection generates a subgroup normal in a Klein-four subgroup that is itself normal in $D_4$ , yet that reflection subgroup is not normal in $D_4$ . Normality is a relationship between a subgroup and its ambient group, never an intrinsic feature of the subgroup by itself. Second, the defining equation $gN=Ng$ is an equality of sets, not of elements: it does not say $gn=ng$ for each $n$ . The coset $gN$ merely coincides with the coset $Ng$ as a whole; individual products may still fail to commute. Reading $gN=Ng$ as elementwise commutation would make every normal subgroup central, which is flatly false — $A_3$ is normal in the non-abelian $S_3$ without being central.

Worked examples

1 · Conjugacy classes and center of the dihedral group D₄ (fully worked)

The symmetry group of the square, $D_4$ , has order 8 with elements $\{e,r,r^2,r^3,s,sr,sr^2,sr^3\}$ where $r^4=e$ , $s^2=e$ , and $srs^{-1}=r^{-1}$ . Find all conjugacy classes, write the class equation, and identify the center.

Step 1. Handle the rotations first. Conjugating a rotation by another rotation does nothing (rotations commute), and conjugating by a reflection inverts it: $s\,r^k\,s^{-1}=r^{-k}$ . So $r$ and $r^3=r^{-1}$ are conjugate, giving the class $\{r,r^3\}$ , while $r^2=r^{-2}$ is conjugate only to itself, giving the singleton $\{r^2\}$ . The identity is its own class $\{e\}$ . Rotation classes: $\{e\}$ , $\{r^2\}$ , $\{r,r^3\}$ .

Step 2. Now the reflections. Compute $r\,s\,r^{-1}=sr^{-2}=sr^2$ , so $s$ and $sr^2$ are conjugate; likewise $r\,(sr)\,r^{-1}=sr^{-1}=sr^3$ , so $sr$ and $sr^3$ are conjugate. The four reflections split into two classes of two: $\{s,sr^2\}$ and $\{sr,sr^3\}$ . This two-way split is special to $D_n$ with $n$ even — the diagonal mirrors and the edge mirrors are geometrically distinct. Reflection classes: $\{s,sr^2\}$ , $\{sr,sr^3\}$ .

Step 3. Assemble the class equation from the five class sizes $1,1,2,2,2$ :

|D_4|=8=1+1+2+2+2.

The two singleton classes are $\{e\}$ and $\{r^2\}$ , so the center is $Z(D_4)=\{e,r^2\}$ , of order 2. This is strictly bigger than $\{e\}$ , exactly as the $p$ -group theorem (with $p=2$ ) demands. The center is $\{e,r^2\}$ , $|Z(D_4)|=2>1$ .

Step 4. Read off the representation-theory payoff: five conjugacy classes means $D_4$ has five inequivalent irreducible representations, and their dimensions $d_k$ satisfy $\sum_k d_k^2=|G|$ , i.e. $1^2+1^2+1^2+1^2+2^2=8$ . Four one-dimensional irreps and one two-dimensional irrep — the character table has exactly five columns, one per class. Five classes ⇒ five irreps, with $4\cdot 1+1\cdot 4=8$ .

2 · Conjugacy classes of S₄ and its normal subgroups

In the symmetric group $S_n$ , two permutations are conjugate if and only if they share the same cycle type. Use this to list the conjugacy classes of $S_4$ (order 24), verify the class sizes sum to $|S_4|$ , and find the nontrivial proper normal subgroups.

Step 1. The cycle types of $S_4$ are the partitions of 4. Count the permutations of each type. The identity $1+1+1+1$ has size 1. Transpositions $2+1+1$ number $\binom{4}{2}=6$ . Double transpositions $2+2$ number $\tfrac{1}{2}\binom{4}{2}=3$ . Three-cycles $3+1$ number $4\cdot 2=8$ (choose the fixed point, then two cyclic orderings). Four-cycles number $3!=6$ . Class sizes: $1,6,3,8,6$ .

Step 2. Check the total against the group order:

1+6+3+8+6=24=|S_4|.

Five classes again, so $S_4$ has five irreducible representations. The census closes: $24=24$ .

Step 3. A normal subgroup must be a union of whole classes that also forms a subgroup (so it must contain $\{e\}$ and have order dividing 24). Test the class unions: $\{e\}$ together with the three double transpositions gives $1+3=4$ elements — the Klein four-group $V_4$ , closed under multiplication, hence normal. Adjoining the eight three-cycles gives $1+3+8=12$ elements — the alternating group $A_4$ , of index 2, hence normal. The nontrivial proper normal subgroups are $V_4$ (order 4) and $A_4$ (order 12).

Step 4. Notice what fails: the transpositions (size 6) plus $\{e\}$ would be size 7, which does not divide 24 and is not closed, so it cannot be normal — indeed it is not even a subgroup. The chain $\{e\}\trianglelefteq V_4\trianglelefteq A_4\trianglelefteq S_4$ is a composition series, and its successive quotients $\mathbb{Z}_2,\mathbb{Z}_3,\mathbb{Z}_2$ are exactly why $S_4$ is solvable — the fact Galois needed to decide solvability of the quartic. $S_4$ is solvable; $S_n$ for $n\ge 5$ is not, because $A_n$ becomes simple.

Physics of the portrait

Conjugacy and normality are not neutral bookkeeping — they are the exact concepts a twenty-year-old invented to answer a two-hundred-year-old question about polynomial equations, and which physics later borrowed wholesale to run symmetry through quantum mechanics.

Origins · 2 minÉvariste Galois and how this was discovered

ÉG

Évariste Galois

1811–1832, French mathematician · the origin of normal subgroups, groups, and solvability

Galois is the improbable source of nearly every idea on this page. Before him "group" was not a mathematical word; the question of the day, unanswered since Cardano, was why the general quintic cannot be solved by radicals when the quadratic, cubic, and quartic can. Galois's answer, worked out as a teenager and set down in a memoir the Académie repeatedly mislaid or rejected, was to attach to each polynomial a finite group of permutations of its roots — the Galois group — and to show that solvability by radicals corresponds to being able to break that group apart through a chain of what he called décompositions propres: the subgroups we now call normal. His criterion is exactly the composition series of Worked Example 2: an equation is solvable by radicals if and only if its group has a chain of normal subgroups with abelian quotients. That is why $S_4$ is solvable (the quartic has a formula) but $S_5$ is not (the general quintic does not), the obstruction being that $A_5$ is simple — it has no nontrivial normal subgroup to quotient by. Galois wrote the decisive letter to his friend Auguste Chevalier on the night of 29 May 1832, urging him to publish the manuscript, and died the next day of wounds from a duel, aged twenty. Liouville understood and printed the work in 1846, and by the time Camille Jordan's 1870 Traité systematized normal subgroups and composition series, group theory had become a subject. Physics inherited it whole: the conjugacy classes label the columns of every character table, and the normal subgroups organize how symmetry groups factor — the inversion subgroup that sorts states by parity, the rotation subgroup of a point group, each a normal subgroup one quotients out to read a spectrum.

History

1832

— Galois's final letter names invariant (normal) subgroups as the key to solvability by radicals

1844

— Cauchy develops the systematic calculus of permutations and their cycle structure

1846

— Liouville edits and publishes Galois's manuscripts, making the theory accessible

1870

— Jordan's Traité des substitutions consolidates normal subgroups and composition series

1872

— Sylow proves the existence theorems for subgroups of prime-power order

1889

— Hölder formalizes the quotient group and, with Jordan, the Jordan–Hölder theorem

Going further

Texts. Hassani, Mathematical Physics, ch. on Group Theory (conjugacy, class equation, quotient groups) · Arfken & Weber, Mathematical Methods for Physicists, 7th ed., ch. on Group Theory · Stone & Goldbart, Mathematics for Physics, ch. on Group Representation Theory · Byron & Fuller, Mathematics of Classical and Quantum Physics, ch. on Group Theory · Hamermesh, Group Theory and Its Application to Physical Problems · Tinkham, Group Theory and Quantum Mechanics · Nakahara, Geometry, Topology and Physics, §4 on Lie groups for the continuous analogue. Key numerical anchors: $S_3$ gives $6=1+2+3$ with $A_3$ normal of index 2; $D_4$ gives $8=1+1+2+2+2$ with $Z(D_4)=\{e,r^2\}$ and $\sum d_k^2=1+1+1+1+4=8$ ; $S_4$ gives $24=1+6+3+8+6$ with normal subgroups $V_4$ and $A_4$ . The $SO(3)$ character $\chi_j(\theta)=\sin((2j+1)\theta/2)/\sin(\theta/2)$ is the standard textbook value.

Prerequisites. Groups, Subgroups, and Cosets · Group Homomorphisms and Quotient Groups. Leads to. Characters and Orthogonality Theorems · Character Tables and Decomposition of Representations · Group Representations · The Rotation Group SO(3).

Ask the tutor: Ask me to build the full character table of $D_4$ or $S_4$ from the conjugacy classes we just found, and watch each class become a column.

What is the difference between the centralizer and the center?

The centralizer $Z_G(g)$ is attached to a single element $g$ : it is the set of all group elements that commute with that one $g$ , and it is always a subgroup containing $g$ itself. The center $Z(G)$ is the set of elements that commute with everything — it is the intersection of all the centralizers, $Z(G)=\bigcap_g Z_G(g)$ . The centralizer measures how big $g$ 's conjugacy class is, through $|C(g)|=[G:Z_G(g)]$ : a large centralizer means a small class. The center collects exactly the elements whose class has size one. So the center is the common ground of every centralizer, and an element is central precisely when its centralizer is the whole group.

Is every subgroup that "looks symmetric" normal?

No — normality has a precise test that intuition often gets wrong. A subgroup $N$ is normal if and only if $gNg^{-1}=N$ for every $g\in G$ , equivalently if $N$ is a union of complete conjugacy classes. A subgroup can be perfectly natural-looking and still fail this: in $S_3$ the two-element subgroup $\{e,(12)\}$ is a legitimate subgroup, but $(12)$ sits in a three-element class, so the subgroup contains only part of a class and is not normal. The reliable procedure is never to eyeball it but to check the class-union criterion, or to check whether the subgroup is the kernel of some homomorphism — kernels are automatically normal. Two shortcuts always work: any index-2 subgroup is normal, and the center is normal.

Why do physicists care about conjugacy classes specifically?

Because characters are class functions. A character is the trace of a representation matrix, $\chi(g)=\operatorname{tr}D(g)$ , and trace is invariant under conjugation: $\chi(aga^{-1})=\operatorname{tr}(D(a)D(g)D(a)^{-1})=\operatorname{tr}D(g)=\chi(g)$ . So a character takes one value on an entire conjugacy class, which is why a character table has one column per class rather than one per element — an enormous compression. A deeper theorem then says the number of inequivalent irreducible representations equals the number of conjugacy classes exactly. Since decomposing physical states (vibrations, orbitals, multiplets) into irreducibles is done entirely with characters, the very first step in any application is to list the classes of the relevant symmetry group.

How is a normal subgroup connected to a conservation law or a quotient?

A normal subgroup is what you are allowed to "divide out". When $N\trianglelefteq G$ , the cosets form the quotient group $G/N$ , and representations of $G$ that act trivially on $N$ are really representations of $G/N$ in disguise. Physically this factorizes degrees of freedom: if the inversion operation generates a normal subgroup, states organize by parity, and the parity label is a representation of the quotient. In crystallography the translations form a normal subgroup of a space group, and the quotient is the point group. The kernel of any symmetry action is normal, so whenever a symmetry acts and something is left invariant, a normal subgroup and its quotient are lurking — the quotient captures the "effective" symmetry after the invariant part is stripped away.

What makes a group simple, and why does it matter for the quintic?

A group is simple when its only normal subgroups are the trivial ones, $\{e\}$ and the whole group — there is nothing nontrivial to quotient by. Simple groups are the atoms of finite group theory: the Jordan–Hölder theorem says every finite group breaks into a canonical stack of simple "composition factors", much as integers factor into primes. This is exactly Galois's criterion for solvability by radicals. An equation is solvable by radicals precisely when its Galois group has a composition series with abelian (equivalently cyclic-prime) factors. For $n\ge 5$ the alternating group $A_n$ is simple and non-abelian, so it appears as a non-abelian composition factor of $S_n$ — an unbreakable, non-abelian atom — which is why the general quintic has no radical formula while the quartic ( $S_4$ , all of whose factors are $\mathbb{Z}_2$ and $\mathbb{Z}_3$ ) does.

Can normality fail to be transitive, and does it ever cause trouble?

Yes, and it is a classic pitfall. " $N$ normal in $H$ and $H$ normal in $G$ " does not imply " $N$ normal in $G$ ". A standard witness lives in $D_4$ : take a reflection subgroup $\langle s\rangle$ of order 2, which is normal inside the Klein-four subgroup $\{e,r^2,s,sr^2\}$ , which is in turn normal in $D_4$ ; yet $\langle s\rangle$ is not normal in $D_4$ , since conjugating $s$ by $r$ produces $sr^2\notin\langle s\rangle$ . The subgroups that are stable under this chaining are called characteristic subgroups (invariant under all automorphisms, not just inner ones), and characteristic-in-normal does imply normal. In practice the lesson is to always re-test normality against the top group $G$ directly and never to assume it propagates up a tower.