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Finite group algebras and the characteristic condition in Maschke's theorem

Example

Let $G$ be a finite group and $k$ a field. In the group Hopf algebra $k G$ , the left and right integral spaces are both generated by $Λ_{G} = \sum_{g \in G} g$ , and $ε (Λ_{G}) = ∣ G ∣ 1_{k}$ . Consequently $k G$ is semisimple exactly when $char (k) = 0$ or $char (k) = p > 0$ does not divide $∣ G ∣$ .

Facts & Assumptions

Given: A finite group $G$ , its group Hopf algebra $k G$ , and a field $k$ .

[F1]

Group algebras, finite-group function algebras, and polynomial Hopf algebras: the group elements form a basis of $k G$ , with $Δ (g) = g \otimes g$ , $ε (g) = 1$ , and multiplication induced by the group product.

[F2]

Left and right integrals and integral functionals: $t$ is a left integral when $h t = ε (h) t$ for all $h$ ; the right-integral equation is $t h = ε (h) t$ .

[F3]

Hopf–Maschke theorem by integral averaging: a finite-dimensional Hopf algebra over a field is semisimple iff it has a left integral with nonzero counit.

Verification

1.1F1F2algebra

For each $h \in G$ , left multiplication permutes the group basis, so $h Λ_{G} = \sum_{g} h g = \sum_{g} g = Λ_{G}$ ; similarly $Λ_{G} h = Λ_{G}$ . Since $ε (h) = 1$ , these identities show $Λ_{G}$ is both a left and right integral.

2.1F1F2step 1.1algebra

Let $t = \sum_{g} a_{g} g$ be a left integral. For any $h, u \in G$ , comparison of the coefficient of $u$ in $h t = t$ gives $a_{h^{- 1} u} = a_{u}$ . For arbitrary $u, v \in G$ , taking $h = u v^{- 1}$ yields $a_{v} = a_{u}$ , so all coefficients are equal and every left integral is a scalar multiple of $Λ_{G}$ . If instead $t = \sum_{g} a_{g} g$ is a right integral, the coefficient of $u$ in $t h = t$ gives $a_{u h^{- 1}} = a_{u}$ for every $h, u \in G$ ; choosing $h = v^{- 1} u$ gives $a_{v} = a_{u}$ for arbitrary $u, v$ . Thus every right integral is also a scalar multiple of $Λ_{G}$ , and both integral spaces are $k Λ_{G}$ .

3.1F3step 2.1algebra∎

Applying the counit gives $ε (Λ_{G}) = \sum_{g} 1_{k} = ∣ G ∣ 1_{k}$ . Hence a left integral has nonzero counit exactly when this field element is nonzero: this always holds in characteristic zero, and in positive characteristic $p$ it holds exactly when $p ∤ ∣ G ∣$ . By [F3], this is equivalent to semisimplicity of $k G$ .

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