## Decomposing the Group Algebra Into the Direct Sum of Matrix Algebras

**Point of post: **In this post we show how the group algebra of a finite group can decomposed into the direct sum of group algebras such that the isomorphism between them is unitary when the group algebra is given the usual inner product and the direct sum of matrix algebras the usual direct sum inner product where each summand is given the Hilbert-Schmidt inner product. Moreover, the isomorphism shall in fact be also an associative unital algebra isomorphism.

*Motivation*

We have seen that the matrix entry functions form an orthonormal basis for the group algebra. From this we got the important result that . It thus follows from basic linear algebra that as complex vector spaces in this post we’ll show much more. Indeed, we’ll show that this isomorphism is also an associative unital algebra isomorphism. Moreover, we’ll even show that if one gives each an inner product which is a multiple of the Hilbert-Schmidt inner product and the direct sum the usual inner product on direct sums that the isomorphism is also a unitary map! Thus, in essence the group algebra in all ways important becomes isomorphic to an object we know much, much about.

*Decomposing the Group Algebra*

For the remainder of the post give every algebra of the form the Hilbert-Schmidt inner product with a scaling factor of and every direct sum of matrix algebras the usual inner product on direct sums. With this in mind:

**Theorem: ***Let be a finite group. Then, the map*

* *

* *

* *

* *

*,where when useful we denote by , given by*

* *

* *

*(we’ve implicitly chosen some choice of representative matrix entry functions, but of course as usual it is independent of choice) where denotes the entry of the coordinate of an element of the codomain is a unitary associative unital algebra isomorphism.*

**Proof: **We first show that is an associative unital algebra homomorphism. Indeed, note that

where we’ve denoted by what we usually call . Thus, we have that so that the fact that multiplicative identities are mapped to each other is true. Next, to see that is multiplicative we note that for any one has

which is the general entry in the coordinate of and so . Lastly, to verify that really is an algebra homomorphism we must prove that it is a linear combination. But, this is clear since for every and one has

Thus, if we can prove that is unitary then it will be injective, but we know that the domain and codomain are of finite and equal dimension and so is an injective if and only if its an isomorphism and so we’ll be done. So, to see that is unitary we note that we only have to prove that for every . Indeed,

Thus, carries the orthonormal basis to an orthonormal basis, and thus is unitary as claimed. The conclusion follows from previous discussion.

**References:**

Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Math. Soc., 1996. Print.

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