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 statement. Let be a finite group. The center of the group algebra , denoted , consists of elements of the form where the coefficients are constant on conjugacy classes. A basis for is given by the set of class sums .

Let be an integer such that . The map given by extends linearly to a map on . Let's call its restriction to the center . Explicitly, for a class sum , This resulting element is indeed in . If , then any conjugate is also in . The image of the class under is . This set is closed under conjugation, since , so it is a union of conjugacy classes. Thus is a central element.

The claim is that is a ring homomorphism, i.e., for any , we have .

Here are two approaches to a more "direct" proof. The first is combinatorial but relies on a difficult identity. The second uses modular arithmetic and is perhaps the most satisfying algebraic proof that avoids character theory.

1. The Combinatorial Approach (Brauer's Proof)

This approach tackles the problem head-on by trying to prove the homomorphism property on the basis of class sums. We need to show that for any two conjugacy classes and :

Let's expand both sides:

• .
• LHS: .
• RHS: .

So, the problem reduces to proving the following identity in the group algebra :

This is not an element-wise identity. Your suggestion of finding such that for individual is not generally true. The identity only holds after summing over all pairs from the conjugacy classes.

To prove this identity, one can show that for any element , its coefficient on the left-hand side equals its coefficient on the right-hand side.

1. Coefficient on LHS: .
2. Coefficient on RHS: .

Since , the map is a permutation of . Let be an integer such that . Then the inverse map is . The condition is equivalent to . So the LHS coefficient is the number of ways to write as a product with . This is a structure constant of the algebra .

For the RHS, we use the fact that the map bijectively maps any conjugacy class onto another conjugacy class, which we can denote . So, as runs through , runs through (bijectively), and as runs through , runs through (bijectively). The map is a bijection from to . Therefore, the RHS coefficient is . This is another structure constant.

The proof thus boils down to showing the equality of structure constants: for all classes and elements .

While this identity is true, proving it by purely combinatorial means (i.e., constructing an explicit bijection) is surprisingly difficult and technical. It is often called Brauer's Permutation Lemma in a generalized form. The standard proof of this identity itself uses character theory! So this path, while "direct", quickly leads back to the machinery we wanted to avoid.

2. The Modular "Lifting" Proof (A High-Concept Explanation)

This is arguably the most elegant proof that avoids character theory. It works by showing the homomorphism property holds modulo every prime (not dividing ), and then "lifts" the result to .

Step 1: Reduce to primes. The maps have the property that . Since any integer with is a product of primes (that don't divide ) and possibly , it suffices to prove that is a ring homomorphism for each such prime , and that is a homomorphism.

• Case n=-1: where . This map is an anti-automorphism of . Since is a commutative ring, any anti-automorphism is also an automorphism. So . This case is easy.

• Case n=p (prime): Let be a prime such that . We want to show that for any , the element is zero.

Step 2: Work modulo p. Let's consider the ring . The center is a commutative algebra over the field . For any element , we have a remarkable identity: Proof of the identity: Let . A famous identity (sometimes called Jacobson's formula) states that in any associative algebra over , is a sum of (Lie) commutators. Since is central, it commutes with all its summands , but the summands do not commute with each other. A more direct argument is needed. The element can be shown to lie in the commutator subspace . Since is central, is central. The element is also central. Their difference, , lies in . Since , the group algebra is semisimple (Maschke's Theorem). For any semisimple algebra over a field, the intersection of the center and the commutator subspace is zero. Thus, .

Step 3: Show the homomorphism property holds modulo p. Let be the reduction-mod- map. For , let and . Since and are in the center , they commute. So . Using our identity again: and . Putting it all together: This means that .

Step 4: Lift the result from mod p to integers. The coefficients of the element in are all divisible by . This argument holds for every prime that does not divide . If , there are infinitely many such primes. An integer that is divisible by infinitely many distinct primes must be zero. Therefore, all coefficients of are zero, which means .

This completes the proof that is a ring homomorphism. By composition, is a ring homomorphism for any coprime to . Finally, since is invertible (with inverse ), it is a ring automorphism.

This second proof is very powerful. It avoids the messy combinatorics of the first approach and the representation theory of the standard one, using instead a fundamental property of algebras in prime characteristic. It beautifully illustrates the technique of proving an identity over by proving it modulo infinitely many primes.