To address the question of showing that the map ( g \mapsto g^n ) (with ( n ) coprime to ( |G| )) induces an algebra automorphism on the center ( \mathbb{Z}[G]^G ) of the group algebra ( \mathbb{Z}[G] ), we proceed as follows:

Key Observations:

1. Center of the Group Algebra: The center ( Z(\mathbb{Z}[G]) ) (denoted ( \mathbb{Z}[G]^G ) in the problem) is generated by the **c...

Yes. Here is a direct, character-free argument, together with the requested elementwise consequence.

1. A purely algebraic proof via the conjugation permutation module

• Let V = ℤ[G] be the free abelian group with basis {e_g | g ∈ G}, equipped with the G-action by conjugation: x·e_g = e_{x g x^{-1}}. Then End_G(V) (endomorphisms commuting with this action) is naturally identified with the algebr...

Let k be a commutative ring in which |G| is invertible (take k = ℤ, ℚ or ℂ if you wish)
and write

H = kG (the group algebra),
Ĥ = Map(G,k) ≅ Homk (H , k) (all k–valued functions on G),
ĤG = { class functions } = k[G]G (functions constant on conjugacy classes).

1. Convolution on functions and the Hopf-dual ...

Excellent question. The fact that the power map (for coprime to ) induces an algebra automorphism on the center of the group algebra is a beautiful and deep result. The standard proof via character theory is slick but, as you note, feels somewhat indirect. You're asking for a proof that works "natively" within the group algebra itself.

Let's first establish the precise st...

Let

  Φₙ : ℤ[G] → ℤ[G],  g ↦ gⁿ     (gcd(n,|G|)=1)

and denote by Z(ℤG)=ℤ[G]^{G} the centre / class algebra.
We want to show that the restriction

  Φₙ : Z(ℤG) → Z(ℤG)

is a ring automorphism, i.e. Φₙ(z₁z₂)=Φₙ(z₁)Φₙ(z₂) for all central elements z₁,z₂.
A completely elementary proof that avoids every trace of character theory seems not to be known; one can, however, give a very short concep...

<think> We are given a finite group G and an integer n coprime to |G|. The map g ↦ g^n is not necessarily a group homomorphism, but it induces a linear automorphism on the center of the group algebra, specifically on the G-invariant functions under convolution. Surprisingly, this induced map is also an algebra automorphism. The question asks for a direct proof of this fact without using character ...

Below is a “purely class‐algebra” proof which never mentions irreducible characters, but only uses two simple facts

1. If n is coprime to |G| then raising to the n-th power is a bijection on each conjugacy class.
2. The centre Z(ℤ[G]) has a ℤ–basis given by the sums of elements in each conjugacy class, and its multiplication is governed by the obvious “class–multiplication” constants.

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Galois Symmetry on Centre of Group Algebra

This is an interesting question about a surprising property of the power map (where ) when restricted to the center of the group algebra.

Understanding the Problem

First, let's clarify what we're looking at:

• We have a finite group and an integer coprime to
• The map isn't a group homom...

This is a beautiful question about a fundamental symmetry in group theory. Let me provide both a conceptual explanation and a direct proof.

Conceptual Explanation

The key insight is that the map (for ) induces an automorphism on conjugacy classes. This is because:

1. If and are conjugate, then and are also conjugate
2. The map on conjugacy cl...