Conjugacy Classes
For , is a conjugate of if for some . Being conjugate is an equivalence relation, so splits into disjoint conjugacy classes , one class per element. The centralizer is the subgroup of elements that commute with — equivalently, that fix under conjugation.
Conjugation by reads from 's point of view: first undoes , then does , then redoes , so it is "the same action as , relative to a different vantage point." Two elements are conjugate exactly when some change of vantage point turns one into the other, which is why conjugate elements always share every property that does not depend on labeling — order, whether the element is the identity, and (Section 2) cycle structure in a symmetric group. Equivalence follows straight from the group axioms: reflexive with , symmetric by conjugating back with , transitive by composing two conjugators — so, exactly like cosets (Topic 5), conjugacy classes partition with nothing left out and nothing counted twice. is always a subgroup (the same closure argument as Topic 3's subgroup tests), and Section 3 shows it controls the SIZE of completely.
Let be the symmetries of a square, with a rotation and a reflection, satisfying and . Compute : since , . So and are conjugate — reflecting the square turns a rotation into its reverse.
Conjugate by every element of . Rotations commute with (the rotation subgroup is abelian), so whenever is a rotation. By the previous example, conjugating by the reflection gives instead, and the identical relation shows every other reflection does too. So and , the rotation subgroup, of order . Already — a preview of Section 3's main theorem.
Conjugating every element of by every other element (the same table lookups as above) sorts the elements into exactly classes: , , , , . The identity is always alone, since for every ; and is alone too, since commutes with every rotation and as well — is a genuinely central element. The sizes already add to , which Section 3 turns into a general counting law.
A class with more than one element does not mean every class has that size. In , the five classes have sizes , not a uniform size — only in an ABELIAN group (where always, so every class is a singleton) does every class collapse to size . Guessing uniform class sizes here would wrongly predict elements per class, not even an integer.
Find the conjugacy class and the centralizer of in .
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Conjugating by : . Conjugating by : using (the relation with replaced by ), . Every other element gives either or again, so , and , of order , matching .
Conjugacy in Symmetric Groups
For , is obtained from 's disjoint cycle decomposition by applying to every entry: if , then . Consequently two permutations of are conjugate if and only if they have the same cycle type (Topic 4): the same multiset of cycle lengths.
Conjugation by is literally relabeling: everywhere moved the point , moves the point instead, so the pattern of the motion — how many cycles, how long each one is — survives untouched while only the names attached to it change. This is the cleanest illustration yet of Section 1's "change of vantage point" idea: IS the relabeling, and is seen through it. The "if" direction (same cycle type conjugate) is just as constructive: line the two cycle decompositions up side by side and read off a valid directly, as the examples below do.
Let and . Relabel every entry of by : , , , and , . So . Composing , , directly as two-row permutations, with no relabeling shortcut, confirms the same answer, .
Are and conjugate in ? Both have cycle type , so yes — and matching the cycles entry by entry names a explicitly: send and , giving one-line . Checking: relabeling by this gives exactly, and direct composition confirms .
How many elements of have the same cycle type as — one transposition, one -cycle, one fixed point? Choose which of the points form the transposition, then which of the remaining form the -cycle (in one of cyclic orders), leaving point fixed: . Equivalently, the general class-size formula divides by the symmetries of the cycle picture itself — ways to write the -cycle, ways to write the -cycle — giving , the same count.
Moving the same NUMBER of points is not the same as sharing a cycle type. and both move all points, but one is a single -cycle and the other is two -cycles — different cycle types, so NOT conjugate. Centralizers confirm it: has order (class size ), while has order (class size ) — different class sizes, so the two permutations cannot be conjugate to each other.
Are and conjugate in ? If so, give a .
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Both are -cycles, so yes. Matching entries in order, , gives , i.e. (with fixed). Relabeling by this gives exactly, confirmed by direct composition.
The Class Equation
: the size of a conjugacy class equals the index of its centralizer (proved in Section 5). Summing over one representative from each class,
the class equation. The center collects exactly the elements with , i.e. the classes of size , so the equation is usually split as
This is Lagrange's theorem (Topic 5) applied once per conjugacy class: each class is in bijection with the cosets of its own centralizer (Section 5), so its size is automatically a divisor of , and the classes — being a partition of — must add up to all of with nothing left over. Splitting off matters because every remaining term is STRICTLY bigger than (a noncentral element's centralizer is a proper subgroup), which is exactly the lever Section 4 uses to pin down from alone, without finding a single actual element.
From Section 1, 's five conjugacy classes have sizes , so the class equation reads . The two size- classes are exactly , so , and the remaining elements split into three classes of noncentral elements.
In (), cycle types sort the elements into counts of identity, transpositions, three-cycles, double-transpositions, and four-cycles, matching (Topic 4). Each count is also a centralizer index: the identity's centralizer is all of (order , class size ); a transposition's centralizer has order (class size ); a -cycle's has order (class size ); a double-transposition's has order (class size ); a -cycle's has order (class size ). Only the identity has class size , so — like every with , has a trivial center.
Could a group of order have conjugacy classes of sizes ? Every class size is a centralizer index, hence a divisor of by Lagrange — but . So no group of order can have a conjugacy class of size at all, and is impossible as a class equation, regardless of which group of order is checked.
The terms of the class equation are INDICES, , not sizes of subgroups sitting inside — a conjugacy class is almost never a subgroup itself. 's transpositions are not even closed: , a -cycle, not a transposition, and the set of transpositions does not even contain the identity.
Find the class equation for and identify .
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By cycle type: identity, three-cycles, transpositions, so the classes have sizes and . Only the identity has class size , so — like every with , has a trivial center.
p-Groups and Their Centers
If for a prime and , then ; in fact divides . A -group is any group whose order is a power of a single prime.
Every noncentral class size divides (Lagrange) and is bigger than , so it is some with — hence a multiple of . The class equation then reduces modulo to , since and every term in the sum are multiples of . always contains , so ; being also a multiple of forces . The whole proof is a counting argument — it never needs to exhibit an actual central element.
, a -group with . From Section 3, , and indeed — consistent with the theorem, and about as small as a nontrivial center can be.
Does every group of order have more than one element commuting with everything? Since , the theorem applies with : divides , so — in particular . Yes, every group of order has a nontrivial center, with no need to construct the group at all.
Does every group of order have to be abelian? By the theorem, divides , so . If , pick any : then (strict since ), so is a divisor of strictly bigger than — forcing , i.e. commutes with everything after all, contradicting . So is impossible, , and every group of order is abelian (full proof in Section 5).
A nontrivial center does not mean abelian — only does. has a nontrivial center , yet itself is not abelian: from the table in Section 1, while , and . A nontrivial center is a strong structural clue, not a proof of commutativity.
Must every group of order have a nontrivial center?
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, so divides by the theorem, giving — in every case . Yes, every group of order has a nontrivial center.
Why It Works
Reflexive: . Symmetric: if then , so is a conjugate of via . Transitive: if and , then , using (Topic 1), so is a conjugate of via . An equivalence relation always partitions its set into disjoint classes, so the conjugacy classes partition .
Let . Map the left cosets of to by . This is well defined and injective in one stroke: for , (the elements fixing under conjugation are exactly , by definition) (Topic 5). So distinct cosets give distinct conjugates and equal cosets give equal conjugates, and the map is onto by definition of the class — a bijection between the cosets of and , giving . Since the conjugacy classes partition , summing over one representative per class recovers exactly once each — the class equation.
Let . For any , : applying to gives exactly of whatever does to . So the arrow becomes the arrow . Applying this to every arrow of a cycle of turns it into the cycle of , of the same length , and since is a bijection, distinct cycles of stay disjoint after relabeling. So has exactly 's cycle type — conjugate permutations share a cycle type. Conversely, suppose and have the same cycle type: write and , with matching lengths listed in the same order, padding with -cycles for fixed points so every point of appears exactly once in each decomposition. Define by , , and so on — a well-defined bijection of , since both decompositions list every point exactly once. By the forward direction, relabels 's cycles by , turning into and so on — exactly . So and are conjugate.
Let . For each noncentral (i.e. ), is a proper subgroup of (proper since some element fails to commute with ), so by Lagrange is a divisor of bigger than — hence for some , and in particular a multiple of . The class equation groups the noncentral elements into classes and reads
Every term on the right of the sum is a multiple of , and is too, so is a difference of multiples of , hence itself a multiple of . Since always, ; combined with , this forces , so .
Let . By the previous proof, is a multiple of dividing (Lagrange), so . Suppose toward a contradiction that , and pick any (nonempty, since ). The centralizer contains (central elements commute with everything, including ) and also contains itself, which is not in — so strictly, giving . But also divides by Lagrange, and the only divisor of exceeding is itself, so . That makes commute with every element of , i.e. — contradicting the choice of . The contradiction rules out , so , i.e. : every element commutes with every other, and is abelian.