Abstract Nonsense

Crushing one theorem at a time

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 G/\mathcal{Z}(G).

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 G and any \alpha\in\widehat{G} one has that d_\alpha\mid \left|G/\mathcal{Z}\left(G\right)\right| (where, as usual, \mathcal{Z}(G) 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 a,b,c\in\mathbb{Z} be such that c\mid b and \displaystyle a^n\mid \frac{b^n}{c^{n-1}} for every n\in\mathbb{N}. Then, \displaystyle a\mid \frac{b}{c}.

Proof: Let p be a prime and let p^x,p^y and p^z are the maximal powers of p dividing a,b and c respectively. By our assumption we clearly have that nx\leqslant ny-(n-1)z for every n\in\mathbb{N}. In particular, choose n=z+1, then one has that (z+1)x\leqslant (z+1)y+z^2 or x\leqslant y-z+\frac{z}{z+1} since x,y,z are integers it follows that x\leqslant y-z and thus in particular p^x\mid p^{y-z}. Since p was an arbitrary prime the conclusion follows. \blacksquare

 

And now, for the big theorem:

The  Dimension Theorem (Strong Version)

Theorem: Let G be a finite group and \alpha\in\widehat{G}. Then, d_\alpha\mid \left|G/\mathcal{Z}\left(G\right)\right|.

Proof: Consider the direct product G^n and \mathcal{Z}\left(G\right)^n\unlhd G^n. Then, for some chosen representative \rho^{(\alpha)}\in \alpha define the representation \psi_n^{(\alpha)}:G^n\to\mathcal{U}\left(\mathscr{V}^{\otimes_n}\right) by

 

\psi_n^{(\alpha)}(g_1,\cdots,g_n)=\rho^{(\alpha)}(g_1)\otimes\cdots\otimes\rho^{(\alpha)}(g_n)

 

We claim that \psi_n^{(\alpha)} is irreducible. Indeed, note that it’s character \chi_{\psi_n^{(\alpha)}} is given by

 

\displaystyle \chi_{\psi_n^{(\alpha)}}(g_1,\cdots,g_n)=\prod_{j=1}^{n}\chi^{(\alpha)}(g_j)

 

Thus,

\displaystyle \begin{aligned}\left\langle \chi_{\psi_n^{(\alpha)}},\chi_{\psi_n^{(\alpha)}}\right\rangle &= \sum_{(g_1,\cdots,g_n)\in G^n}\chi_{\psi_n^{(\alpha)}}(g_1,\cdots,g_n)\overline{\chi_{\psi_n^{(\alpha)}(g_1,\cdots,g_n)}}\\ &= \sum_{(g_1,\cdots,g_n)\in G_n}\prod_{j=1}^{n}\chi^{(\alpha)}(g_j)\overline{\chi^{(\alpha)}(g_j)}\\ &= \prod_{j=1}^{n}\sum_{g_j\in G}\chi^{(\alpha)}\chi^{(\alpha)}(g_j)\overline{\chi^{(\alpha)}(g_j)}\\ &= \prod_{j=1}^{n}1\\ &= 1\end{aligned}

 

and so by previous theorem we have that \psi_n^{(\alpha)} is irreducible. Let z\in\mathcal{Z}(G). One clearly has that \rho^{(\alpha)}(z)\rho^{(\alpha)}(g)=\rho^{(\alpha)}(g)\rho^{(\alpha)}(z) and thus by Schur’s lemma one has that \rho^{(\alpha)}(z)=f(z)\mathbf{1}. Moreover, it’s clear that for any z,z'\in\mathcal{Z}(G) one has that

 

\begin{aligned}f(zz')\mathbf{1} &=\rho^{(\alpha)}(zz')\\ &=\rho^{(\alpha)}(z)\rho^{(\alpha)}(z')\\ &=f(z)\mathbf{1}f(z')\mathbf{1}\\ &=(f(z)f(z'))\mathbf{1}\end{aligned}

 

and so f is a homomorphism(a general reasoning for why this ‘restriction’ to \mathcal{Z}(G) works will be explained when we discuss representations relationship with subgroups). We now define H_n=\left\{(z_1,\cdots,z_n)\in\mathcal{Z}(G)^n:g_1\cdots g_n=e\right\}. Clearly H_n\unlhd G^n since H_n\leqslant \mathcal{Z}(G)^n. Thus, it makes sense to consider G^n/H_n. So, define \omega_n:G^n/H_n\to\mathcal{U}\left(\mathscr{V}^{\otimes_n}\right) by \omega_n\left((g_1,\cdots,g_n)H_n\right)=\psi_n^{(\alpha)}(g_1,\cdots,g_n). To see that this is well-defined we merely note that for any (z_1,\cdots,z_n)\in H_n one has that

 

\displaystyle \begin{aligned}\psi_n^{(\alpha)}(z_1,\cdots,z_n) &= \rho^{(\alpha)}(g_1)\otimes\cdots\otimes \rho^{(\alpha)}(g_n)\\ &= f(z_1)\mathbf{1}_{\mathscr{V}}\otimes f(z_n)\mathbf{1}_{\mathscr{V}}\\ &= \left(f(z_1)\cdots f(z_n)\right)\left(\mathbf{1}_{\mathscr{V}}\otimes\cdots\otimes\mathbf{1}_{\mathscr{V}}\right)\\ &= \left(f(z_1)\cdots f(z_n)\right)\mathbf{1}_{\mathscr{V}^{\otimes_n}}\\ &= f(z_1\cdots z_n)\mathbf{1}_{\mathscr{V}^{\otimes_n}}\\ &= f(e)\mathbf{1}_{\mathscr{V}^{\otimes_n}}\\ &= \mathbf{1}_{\mathscr{V}^{\otimes_n}}\end{aligned}

 

from where well-definedness almost immediately follows. We now claim that \omega_n is an irrep. To see this, suppose that \mathscr{W}\leqslant\mathscr{V}^{\otimes_n} is \omega_n-invariant. Then, for every (g_1,\cdots,g_n)\in G^n one has that

 

\left(\psi_n^{(\alpha)}(g_1,\cdots,g_n)\right)\left(\mathscr{W}\right)=\left(\omega_n\left((g_1,\cdots,g_n)H_n\right)\right)\left(\mathscr{W}\right)\subseteq\mathscr{W}

 

and thus \mathscr{W} is \psi_n^{(\alpha)}-invariant and thus by the irreducibility of \psi_n^{(\alpha)} it follows that \mathscr{W}=\{\bold{0}\} or \mathscr{W}=\mathscr{V}^{\otimes_n}. Thus, since \mathscr{W} was arbitrary it follows that \omega_n is irreducible. It follows then by the weak version of the dimension theorem that d_\alpha^n\mid \left|G^n/H_n\right|. But, it’s evident that \left|H_n\right|=\mathcal{Z}(G)^{n-1} since the first n-1 coordinates can be chosen freely from \mathcal{Z}(G) and the last coordinate is determined since g_n=(g_1\cdots g_{n-1})^{-1}. Thus, we’ve just proven that

 

\displaystyle d_\alpha^n\mid \frac{|G|^n}{\left|\mathcal{Z}(G)\right|^{n-1}}

 

and since n\in\mathbb{N} was arbitrary the conclusion follows from the lemma. \blacksquare

 

References:

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

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March 7, 2011 - Posted by | Algebra, Representation Theory | , , , , ,

1 Comment »

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

    Pingback by Representation Theory: Character Tables « Abstract Nonsense | March 21, 2011 | Reply


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