## The Dimension Theorem (Strong Version)

**Point of post: **In this post we derive a result which is more precise than that of the dimension theorem. Namely, that the degree of any irrep divides the order of .

*Motivation*

In our last post we proved the incredible fact that the degree of any irrep divides the order of the group. We now take this one step further and show that for any finite group and any one has that (where, as usual, denotes the center of the group). This proof is quite long, but the result is interesting enough to merit the effort. First we need a simple number-theoretic lemma:

**Lemma: ***Let be such that and for every . Then, .*

**Proof: **Let be a prime and let and are the maximal powers of dividing and respectively. By our assumption we clearly have that for every . In particular, choose , then one has that or since are integers it follows that and thus in particular . Since was an arbitrary prime the conclusion follows.

And now, for the big theorem:

*The Dimension Theorem (Strong Version)*

**Theorem: ***Let be a finite group and . Then, .*

**Proof: **Consider the direct product and . Then, for some chosen representative define the representation by

We claim that is irreducible. Indeed, note that it’s character is given by

Thus,

and so by previous theorem we have that is irreducible. Let . One clearly has that and thus by Schur’s lemma one has that . Moreover, it’s clear that for any one has that

and so is a homomorphism(a general reasoning for why this ‘restriction’ to works will be explained when we discuss representations relationship with subgroups). We now define . Clearly since . Thus, it makes sense to consider . So, define by . To see that this is well-defined we merely note that for any one has that

from where well-definedness almost immediately follows. We now claim that is an irrep. To see this, suppose that is -invariant. Then, for every one has that

and thus is -invariant and thus by the irreducibility of it follows that or . Thus, since was arbitrary it follows that is irreducible. It follows then by the weak version of the dimension theorem that . But, it’s evident that since the first coordinates can be chosen freely from and the last coordinate is determined since . Thus, we’ve just proven that

and since was arbitrary the conclusion follows from the lemma.

**References:**

1.Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Mathematical Society, 1996. Print.

[…] Proof: This is just a restatement of the strong version of the dimension theorem. […]

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