Let \(C_p\) denote the cyclic group of order \(p\).
Show that \(C_p\) has two irreducible representations over \(\mathbb Q\) (up to isomorphism), one of dimension 1 and one of dimension \(p-1\).
Let \(G\) be a finite group, and let \(\rho:G\rightarrow \GL_n(\mathbb Q)\) be a representation of \(G\) over \(\mathbb Q\). Let \(\rho_{\mathbb C}:G\rightarrow\GL_n(\mathbb C)\) denote \(\rho\) followed by the inclusion \(\GL_n(\mathbb Q)\rightarrow \GL_n(\mathbb C)\). Thus \(\rho_{\mathbb C}\) is a representation of \(G\) over \(\mathbb C\), called the complexification of \(\rho\). We say that an irreducible representation \(\rho\) of \(G\) is absolutely irreducible if its complexification remains irreducible over \(\mathbb C\).\ Now suppose \(G\) is abelian and that every representation of \(G\) over \(\mathbb Q\) is absolutely irreducible. Show that \(G\cong(C_2)^k\) for some \(k\) (i.e., is a product of cyclic groups of order 2).
Solution
1. As a \(\mathbb Q\)-algebra,\[ \mathbb Q[C_p]\cong \mathbb Q[x]/(x^p-1) \cong \mathbb Q[x]/(x-1)\times \mathbb Q[x]/(\Phi_p(x)). \]
If \(g\) generates \(C_p\), the map \(\mathbb Q[x]\to\mathbb Q[C_p]\) sending \(x\mapsto g\) has kernel \((x^p-1)\). Since \[ x^p-1=(x-1)\Phi_p(x) \] and the two factors are coprime over \(\mathbb Q\), the Chinese remainder theorem gives the product decomposition. The cyclotomic polynomial \(\Phi_p\) is irreducible over \(\mathbb Q\), so \[ \mathbb Q[x]/(\Phi_p(x))\cong\mathbb Q(\zeta_p). \]
2. Hence \(C_p\) has exactly two irreducible rational representations up to isomorphism: the trivial representation of dimension \(1\), and one representation of dimension \(p-1\).
The simple modules over a finite product of fields are exactly the simple modules coming from one factor. Thus the two simple \(\mathbb Q[C_p]\)-modules are \[ \mathbb Q \quad\text{and}\quad \mathbb Q(\zeta_p). \] Their dimensions over \(\mathbb Q\) are \(1\) and \[ [\mathbb Q(\zeta_p):\mathbb Q]=\varphi(p)=p-1. \]
3. If \(p>2\), the \((p-1)\)-dimensional rational irreducible representation of \(C_p\) is not absolutely irreducible.
After extending scalars to \(\mathbb C\), \[ \mathbb Q(\zeta_p)\otimes_{\mathbb Q}\mathbb C \cong \bigoplus_{j=1}^{p-1}\mathbb C_{\chi_j}, \] where \(\chi_j(g)=\zeta_p^j\). Thus its complexification is a direct sum of \(p-1>1\) one-dimensional characters.
4. The nontrivial two-dimensional rational irreducible representation of \(C_4\) is not absolutely irreducible.
We have \[ \mathbb Q[C_4]\cong \mathbb Q[x]/(x^4-1), \] and the factor \(x^2+1\) gives the simple module \[ \mathbb Q(i), \] of dimension \(2\). After complexification, \[ \mathbb Q(i)\otimes_{\mathbb Q}\mathbb C \cong\mathbb C\oplus\mathbb C, \] corresponding to the two characters sending a generator of \(C_4\) to \(i\) and \(-i\). Hence this rational irreducible is not absolutely irreducible.
5. Under the hypothesis in the problem, every element of the finite abelian group \(G\) has order dividing \(2\).
Suppose instead that \(G\) has an element of order greater than \(2\). By the structure theorem for finite abelian groups, one of the following holds:
- some odd prime \(p\) divides the order of an element of \(G\), in which case \(G\) has a quotient isomorphic to \(C_p\); or
- the exponent of \(G\) is divisible by \(4\), in which case \(G\) has a quotient isomorphic to \(C_4\).
4. Inflation preserves irreducibility because a subspace is \(G\)-stable exactly when it is \(H\)-stable, and its complexification remains reducible because the action still factors through \(H\). This contradicts the hypothesis that every irreducible rational representation of \(G\) is absolutely irreducible.
6. Therefore\[ G\cong(C_2)^k \]for some \(k\ge0\).
By 5, every nonidentity element of the finite abelian group \(G\) has order \(2\). Thus \(G\) is naturally a vector space over \(\mathbb F_2\). Since it is finite, it has finite dimension \(k\), so as an abelian group \[ G\cong (C_2)^k. \]