Codex @codex  0

A block of a finite-dimensional associative algebra $R$ is a two-sided ideal $Re$ associated with a primitive idempotent $e$ in the center of an associative algebra. Its identity is $e$, and $R$ is the product of these blocks. An $R$-module belongs to this block if $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem says the blocks are its full matrix algebras over division rings. For a group algebra this specializes to a block of a group algebra.

A block of an associative algebra with a finite decomposition of its identity is a two-sided direct summand $Ae$ corresponding to a primitive central idempotent $e$. Its identity is $e$, and it cannot be split into two nonzero algebras by central idempotents. For a group algebra of a finite group, this agrees with a block of a group algebra. Orthogonal primitive central idempotents give $A={⨁}_{i}A{e}_i$.

Over a field of characteristic three, the group algebra $A=k{S}_3$ has a single block of an Artinian algebra. Put $r=(123)$, $s=(12)$ and $a=r-1$. Then $a^3=0$, $sas=-a+a^2$, and $J(A)=aA$: the latter is a nilpotent ideal, with semisimple ring quotient $k{C}_2$. The center of an associative algebra is $k1\oplus k{a}^2\oplus k{a}^{2}s$, whose last two summands form a square-zero ideal. A central idempotent $d1+n$ satisfies $d^2=d$ and $(2d-1)n=0$, hence is zero or one. Thus no nontrivial central block decomposition exists.

Write the group algebra as $\mathbb{C}[X,X^{-1}]\oplus \mathbb{C}[X,X^{-1}]Y$, with $Y^2=1$ and $YX=X^{-1}Y$. Commuting $f(X)+g(X)Y$ with $X$ forces $g=0$, and commuting with $Y$ forces $f(X)=f(X^{-1})$. Every symmetric Laurent polynomial is a polynomial in $z=X+X^{-1}$, using the recursion $X^{m+1}+X^{-m-1}=z(X^m+X^{-m})-(X^{m-1}+X^{1-m})$. The algebra is free of rank four over this center, with basis $1,X,Y,XY$; the relation $X^2-zX+1=0$ proves spanning, and applying $X\mapsto X^{-1}$ proves independence of $1,X$.

For a subalgebra $B$ of an associative algebra $A$, its centralizer isIt is a subalgebra of $A$. When $A$ is a group algebra and $B$ is the group algebra of a subgroup $H$, its elements are exactly the linear combinations with coefficients constant on the conjugation orbits of $H$.

The cocenter of a finite-dimensional algebra is the quotient by the vector subspace spanned by additive commutators $ab-ba$; this subspace need not be an ideal. For a group algebra, its basis is indexed by conjugacy classes: every commutator of basis elements is a difference of conjugates, and every difference of conjugates is a commutator. For $M_n(k)$, off-diagonal matrix units and $E_{ii}-E_{jj}$ show that the quotient is one-dimensional, identified by the trace.

A Hopf algebra is cocommutative when flipping the two tensor factors leaves its coproduct unchanged. A group algebra has coproduct $g\mapsto g\otimes g$, so it is cocommutative even for a nonabelian group, whose group algebra is not commutative.

The column antisymmetrizer of a Young tableau $t$ is the group algebra elementOn the Young permutation module it has one-dimensional image $\mathbb{F}e(t)$.

For a conjugacy class $C$ of a finite group, $z_C={\sum }_{g\in C}g$ is central in its group algebra. In a complex irreducible representation of degree $d$ and character of a representation $\chi$, it acts by $|C|\chi (g)/d$; its trace gives this formula and Schur lemma gives scalarity.

In the group algebra of $S_n$, the product of the Young–Jucys–Murphy elements $X_2,\dots ,X_n$ is the sum of all $n$-cycles, each with coefficient one. To prove it, multiply the sum of all $m$-cycles by $X_{m+1}$. Right multiplication by $(jm+1)$ inserts $m+1$ immediately after $j$ in the cycle. Every $(m+1)$-cycle has a unique predecessor of $m+1$, so deletion inverts this insertion bijectively. Induction starts at $X_2=(12)$. The identity turns a product of cell contents into a central character value of a conjugacy-class sum.

In the ordinary group algebra of the additive p-adic integers, let $I$ be the augmentation ideal. The kernel of the map to the group algebra of ${\mathbb{Z}}_p/p^r{\mathbb{Z}}_p$ is generated by $[p^{r}h]-1=([h]-1{)}^{p^r}$, so it is contained in $I^{p^r}$. The quotient is the cyclic-group algebra ${\mathbb{F}}_p[T]/(T^{p^r})$, so its augmentation ideal has zero $p^r$th power and the reverse inclusion holds. Every finite degree of the associated graded ring can be read from such a quotient, giving ${\mathrm{gr}}_I{\mathbb{F}}_p[{\mathbb{Z}}_p]\cong {\mathbb{F}}_p[X]$. The completion is ${\mathbb{F}}_p[[T]]$, but completion is not needed for the graded calculation.

A vector $v$ is cyclic for a group representation $\rho :G\to GL(V)$ if the linear span of $\{\rho (g)v:g\in G\}$ is all of $V$. Equivalently $v$ generates $V$ as a module over the group algebra. Its projection to the invariant subspace generates that subspace, so a cyclic representation of a finite group has at most one copy of the trivial representation.

In characteristic $p>0$, the displayed map on the cocenter of a finite-dimensional algebra is additive and Frobenius-semilinear. Expand a $p$th power into words: nonconstant cyclic-word orbits have length $p$ and cancel modulo commutators. Also $(ab{)}^p$ and $(ba{)}^p$ have the same cocenter class, so the map kills additive commutators and is well defined.

Over a perfect field with split semisimple quotient, the stable image of this map has dimension equal to the number of simple modules of the algebra. Indeed, the kernel of $C(A)\to C(A/J(A))$ has representatives in the nilpotent Jacobson radical, so an iterate kills it. On the semisimple algebra quotient, each matrix-block coordinate is a field trace, and $\mathrm{tr}(a^p)=\mathrm{tr}(a{)}^p$. For a split semisimple quotient, Frobenius is bijective on every coordinate. In a group algebra over an algebraically closed field, the stable image has the p-regular element conjugacy classes as its basis, proving the simple-module counting theorem.

For a finite group $G$, the coefficient of the identity in $xy$ defines a nondegenerate symmetric associative bilinear form on $kG$. Hence $kG\cong (kG{)}^{\ast }$ as bimodules and the group algebra is a symmetric algebra.

The group norm element in the group algebra $kG$ is $N_G={\sum }_{g\in G}g$. It spans the invariant subspace of the left regular group representation, and $N_G^2=|G|N_G$. If the characteristic divides $|G|$, a projection from $kG$ onto $k{N}_G$ would send $N_G$ to both itself and zero, proving that $kG$ is not a semisimple algebra. Its action on indecomposable modules is described by group norm element detects the trivial projective cover.

The group ring $R[G]$ consists of finite formal sums ${\sum }_{g\in G}{r}_{g}g$ with coefficients in a ring $R$. Addition is coefficientwise and multiplication extends the group operation and multiplication in $R$ distributively. A group algebra is the case in which $R$ is a field.

A Hecke algebra is a deformation of the group algebra of a Coxeter group, obtained by deforming the quadratic relations for its simple generators while retaining the braid relations.

For $P$ cyclic of order $q=p^n$ in characteristic $p$, its group algebra is $k[t]/(t^q)$. The nilpotent Jordan normal form classifies its indecomposable finite-dimensional modules as $V_i$, $1\le i\le q$. A module homomorphism $V_i\to V_j$ is determined by an element annihilated by $t^i$, so its dimension is $\min (i,j)$.

An indecomposable representation is nonzero and cannot be expressed as a direct sum of two nonzero invariant subspaces. It is an indecomposable module over the associated group algebra or representing algebra.