Group Algebras

The group algebra $k[G]$ has a basis of the group's elements and the group's own multiplication as its product, extended bilinearly. It is the most elementary construction in this section — which is why it comes last rather than first. The interesting part is not the product. It is the centre.

$k[G]$ is commutative exactly when $G$ is, which is usually false. But its centre is always commutative, and it always has a basis of class sums: one element per conjugacy class, obtained by adding that class up. So a non-commutative algebra carries a canonical commutative subalgebra, whose dimension is the number of conjugacy classes — which is also the number of irreducible characters, and the doorway to representation theory.

#Commutative or not

Cyclic groups are abelian; dihedral ones past n = 2 are not.

In$\mathrm{GroupIsAbelian}(\mathrm{CyclicGroup}(6))$Out$\mathrm{True}$In$\mathrm{GroupIsAbelian}(\mathrm{DihedralGroup}(4))$Out$\mathrm{False}$In$\mathrm{GroupOrder}(\mathrm{DihedralGroup}(4))$Out$8$

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The class sums span the centre

#A commutative subalgebra of a non-commutative algebra

A single reflection is not central; the class sum containing it is. The first class is always the identity alone.

In$\mathrm{ClassSum}(\mathrm{DihedralGroup}(3), 2)$Out$\mathrm{GroupBasis}(\text{2})+\mathrm{GroupBasis}(\text{1})$In$\mathrm{IsCentral}(\mathrm{DihedralGroup}(3), \mathrm{ClassSum}(\mathrm{DihedralGroup}(3), 2))$Out$\mathrm{True}$In$\mathrm{IsCentral}(\mathrm{DihedralGroup}(3), \mathrm{GroupBasis}(\text{s0}))$Out$\mathrm{False}$

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<notatio-cell value="ClassSum(DihedralGroup(3), 2)" data-allow-mismatch="children" /> <notatio-cell value="IsCentral(DihedralGroup(3), ClassSum(DihedralGroup(3), 2))" data-allow-mismatch="children" /> <notatio-cell value='IsCentral(DihedralGroup(3), GroupBasis("s0"))' data-allow-mismatch="children" />

The class sums have disjoint supports, so they are linearly independent, so counting them counts the centre's dimension. The tests check that against the closed forms: $n$ classes for $\mathbb{Z}_n$, $(n+3)/2$ for $D_n$ with $n$ odd, $(n+6)/2$ for $n$ even.

#Counting classes

Z₆ is abelian so every element is its own class. D₃ ≅ S₃ has three; D₄ has five.

In$\mathrm{GroupCentreDimension}(\mathrm{CyclicGroup}(6))$Out$6$In$\mathrm{GroupCentreDimension}(\mathrm{DihedralGroup}(3))$Out$3$In$\mathrm{GroupCentreDimension}(\mathrm{DihedralGroup}(4))$Out$5$In$\mathrm{ConjugacyClasses}(\mathrm{DihedralGroup}(3))$Out$\bigl\lbrack\bigl\lbrack\mathrm{GroupBasis}(\text{0})\bigr\rbrack, \bigl\lbrack\mathrm{GroupBasis}(\text{1}), \mathrm{GroupBasis}(\text{2})\bigr\rbrack, \bigl\lbrack\mathrm{GroupBasis}(\text{s0}), \mathrm{GroupBasis}(\text{s1}), \mathrm{GroupBasis}(\text{s2})\bigr\rbrack\bigr\rbrack$

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<notatio-cell value="GroupCentreDimension(CyclicGroup(6))" data-allow-mismatch="children" /> <notatio-cell value="GroupCentreDimension(DihedralGroup(3))" data-allow-mismatch="children" /> <notatio-cell value="GroupCentreDimension(DihedralGroup(4))" data-allow-mismatch="children" /> <notatio-cell value="ConjugacyClasses(DihedralGroup(3))" data-allow-mismatch="children" />

k[Zₙ] is k[x]/(xⁿ−1)

For a cyclic group the product just adds indices mod $n$ — so $k[\mathbb{Z}_n]$ is a polynomial ring modulo a single relation. When the base field has $n$-th roots of unity it splits into $n$ copies of the field, which is the same CRT-flavoured splitting the finite hypercomplex page and the residue numerals keep running into.

#Adding indices

2 + 5 = 7 ≡ 1. And (1 + x)(1 + x⁵) = 2 + x + x⁵.

In$\mathrm{GroupProduct}(\mathrm{CyclicGroup}(6), \mathrm{GroupBasis}(\text{2}), \mathrm{GroupBasis}(\text{5}))$Out$\mathrm{GroupBasis}(\text{1})$In$\mathrm{GroupOrder}(\mathrm{GroupDirectProduct}(\mathrm{CyclicGroup}(2), \mathrm{CyclicGroup}(3)))$Out$6$

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<notatio-cell value='GroupProduct(CyclicGroup(6), GroupBasis("2"), GroupBasis("5"))' data-allow-mismatch="children" /> <notatio-cell value="GroupOrder(GroupDirectProduct(CyclicGroup(2), CyclicGroup(3)))" data-allow-mismatch="children" />

The dihedral relations

Elements are written k for $r^k$ and s k for $s r^k$, with $r^n = 1$, $s^2 = 1$ and $s r s = r^{-1}$.

#s² = 1 and s r s = r⁻¹

s² is the identity, and conjugating r by s inverts it — r³ in D₄.

In$\mathrm{GroupProduct}(\mathrm{DihedralGroup}(4), \mathrm{GroupBasis}(\text{s0}), \mathrm{GroupBasis}(\text{s0}))$Out$\mathrm{GroupBasis}(\text{0})$In$\mathrm{GroupProduct}(\mathrm{DihedralGroup}(4), \mathrm{GroupProduct}(\mathrm{DihedralGroup}(4), \mathrm{GroupBasis}(\text{s0}), \mathrm{GroupBasis}(\text{1})), \mathrm{GroupBasis}(\text{s0}))$Out$\mathrm{GroupBasis}(\text{3})$

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<notatio-cell value='GroupProduct(DihedralGroup(4), GroupBasis("s0"), GroupBasis("s0"))' data-allow-mismatch="children" /> <notatio-cell value='GroupProduct(DihedralGroup(4), GroupProduct( DihedralGroup(4), GroupBasis("s0"), GroupBasis("1"), ), GroupBasis("s0"))' data-allow-mismatch="children" />

Things worth knowing

The group axioms are verified, not assumed. The multiplication tables are checked for associativity, an identity, inverses, and the Latin-square property on every group here — because everything downstream rests on them being groups at all.

Several routes to the same algebra. $k[S_n]$ is also the symmetric-group diagram algebra, and also $H_n(q)$ at $q = 1$. Three libraries, one object.

GroupProduct takes its group, since a basis element carries no reference to the group it came from — the same shape as QuiverCompose.

Not built yet. Character theory: the character table, orthogonality relations, and the decomposition of $k[G]$ into matrix blocks. That is what the centre is the doorway to, and it is the obvious next step. Also only cyclic, dihedral and direct products — permutation groups come in through the diagram and Hecke libraries instead.