Character determines representation in characteristic zero

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Suppose is a finite group and is a field of characteristic zero. Then, the character of any finite-dimensional representation of over completely determines the representation, i.e., no two inequivalent finite-dimensional representations can have the same character.

Note that does not need to be a splitting field.

Related facts

Opposite facts


Facts used

  1. Character orthogonality theorem
  2. Maschke's averaging lemma
  3. Orthogonal projection formula


Given: A group , two linear representations of with the same character over a field of characteristic zero.

To prove: and are equivalent as linear representations.

Proof: By Fact (2), both and are completely reducible, and are expressible as sums of irreducible representations. Suppose is a collection of distinct irreducible representations obtained as the union of all the representations occurring in a decomposition of into irreducible representations and a decomposition of into irreducible representations. In other words, there are nonnegative integers such that:


Let denote the character of and denote by the value (note: this would be 1 if were a splitting field, and in general it is the sum of squares of multiplicities of irreducible constituents over a splitting field).

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 and for each Fact (3) Direct application of fact
2 and for each has characteristic zero, so the manipulation makes sense Step (1)
3 for each has characteristic zero Step (2) [SHOW MORE]
4 and are equivalent Step (3)