In this post, we will give a brief introduction to the representation theory of general (finite) groups.

**Definition: **A representation of a group is a homomorphism , where is called the dimension of the representation.

A representation is faithful if the map is injective. A representation can also be viewed as the action of on an -dimensional -vector space because is isomorphic to via the isomorphism . However, this requires fixing a basis. This leads to the following definition.

**Definition: **Let be a -vector space and a group. A representation of on is a map . This is equivalent to an operation defined via .

An isomorphism of representations and of on and respectively is an isomorphism that is consistent with the action of on and . We will henceforth denote by . If is a representation of , the character of is -valued function on defined as . We claim that characters are constant on conjugate classes. Suppose that . Then . Since , and are conjugate, they have the same trace.

Consider a character associated to a representation . Then is the dimension of the representation of since is the identity on , whose trace is the dimension of . Another important statement is the following. Suppose , and suppose . Then the roots of the characteristic polynomial of are powers of the th roots of unity. Since the trace of a matrix is a sum of its eigenvalues, and if is an eigenvalue of , then is an eigenvalue of , and hence must be a power of . This also shows that .

Let act on a -vector space . A subspace of is -invariant if the action of on reduces to an action of on . If , where and are -invariant subspaces, then the representation on is the direct sum of the representations on and . If is the representation on , then the matrix associated to will be in block diagonal form. This leads to the following definition.

**Definition: **A representation of on a -vector space is irreducible if there is no proper -invariant subspace of .

Let be a non-irreducible (reducible) representation of on , and consider a proper -invariant subspace . The basis of can be extended to , and then the matrix associated to will be in block form, where the upper left matrix acts on the basis of , and the lower right matrix acts on some other representation of . Maschke’s theorem says:

**Theorem: **If is a finite group, every representation of on a nonzero finite-dimensional -vector space is the direct sum of irreducible representations of .

*Proof: *The proof is two-part. Say that a representation on a -vector space is unitary if for all , the matrix associated to is unitary, i.e., . Suppose is a reducible unitary representation of on a Hermitian space . If is a -invariant subspace of , then is also -invariant, because since is unitary, if , then ; since is -invariant, . Since , it follows that is the direct sum of the restrictions to and . Therefore it follows that every unitary representation on a Hermitian space is an orthogonal sum of irreducible representations.

Now it suffices to prove that if is a representation of a finite group on a -vector space , then there is a -invariant positive-definite Hermitian form on (making it a Hermitian space). Consider some positive-definite Hermitian form on , and define

Since is finite, this is clearly -invariant. Checking that this new form is positive-definite and Hermitian is easy, and this proves the required result. QED.

We have Schur’s lemma:

**Theorem: **If are irreducible representations of on and , and if is a -invariant linear transformation, then either is an isomorphism or . In addition, if is a -invariant linear transformation then for some scalar .

*Proof: *To prove the first statement, consider the subspace . Clearly this is -invariant, so is either or . If , then cannot be , so , i.e., is injective. In this case, because is also -invariant, it is either or . Since , the image of is , and is an isomorphism. To prove the second statement, let be an eigenvalue of . Then is not zero since it contains an eigenvector of . Therefore, is not an isomorphism, and so , i.e., . QED.

There are a number of important consequences. For example, suppose that is abelian, and consider an irreducible representation . Because is abelian, , and therefore, for every , by Schur’s lemma, must be scalar multiplication. In this case, if is not one-dimensional, then every subspace of is a -invariant subspace – but this contradicts irreducibility! Therefore, every irreducible representation of an abelian group is one-dimensional. We will see another proof below of this when the abelian group is finite.

**Definition: **Let be characters of . There is a Hermitian product defined via

Because is constant on conjugacy classes it is clear that if denotes the conjugacy classes of and are representatives of , then

The main theorem of representation theory (without proof) is the following.

**Theorem: **Suppose is a finite group. Then:

- The irreducible characters of form an orthonormal basis of class functions (functions that are constant on conjugacy classes) of . If are irreducible representations of , then if and are isomorphic, and is otherwise.
- The number of isomorphism classes of irreducible representations is finite, and is the equal to the number of conjugacy classes of the group.
- Suppose represent the irreducible representations of . Then each divides , and: .

There are many corollaries of this. For example, we get another proof that representations of abelian groups are one-dimensional, in the case when the group is finite. If is finite abelian and is an irreducible character of , then must be one-dimensional. To see this, note that since is finite abelian, the number of conjugacy classes of is , which is also the number of irreducible characters of . Since , each must be . We will conclude with the following result.

**Corollary: **Representations of a finite group are isomorphic if and only if their characters are equal.

*Proof. *Since any representation can be written as , so the associated character can be written as . Clearly by the orthogonal projection formula, completing the proof.

Best,

SKD