Fall 2021

• Idea: take a semidirect product involving \(C_9\) and \(C_7\). We’ll need some facts: \(\mathop{\mathrm{Hom}}(C_m, C_n) \cong C_d\) where \(d = \gcd(m, n)\), and \(\mathop{\mathrm{Aut}}(C_m)\cong C_m^{\times}\) which has order \(\phi(m)\) (since one needs to send generators to generators), which can be explicitly calculated based on the prime factorization of \(m\).

• Some calculations we’ll need:

  • \(\mathop{\mathrm{Aut}}(C_9) \cong C_9^{\times}\cong C_{\phi(9)} \cong C_6\), using that \(\phi(p^k) = p^{k-1}(p-1)\).
  • \(\mathop{\mathrm{Aut}}(C_7) \cong C_7^{\times}\cong C_{\phi(7)}\cong C_6\) using that \(\phi(p) = p-1\).

• To get a nonabelian group, we need a nontrivial semidirect product, so look at \(\mathop{\mathrm{Hom}}(G, \mathop{\mathrm{Aut}}(H))\) in the two possible combinations.

  • \(\mathop{\mathrm{Hom}}(C_7, \mathop{\mathrm{Aut}}(C_9)) \cong \mathop{\mathrm{Hom}}(C_7, C_6) \cong C_1 \coloneqq\left\{{e}\right\}\) using that \(\mathop{\mathrm{Hom}}(C_m, C_n) \cong C_{d}\) for \(d = \gcd(m, n)\). So there are no nontrivial homs here, so only the direct product is possible.
  • \(\mathop{\mathrm{Hom}}(C_9, \mathop{\mathrm{Aut}}(C_7)) \cong \mathop{\mathrm{Hom}}(C_9, C_6) \cong C_3\), so use this!
  • Note that we don’t have to consider possibilities for \(C_3\times C_3\), since including this as a factor would yield no elements of order 9.

• So take \(G\coloneqq C_7 \rtimes_\psi C_9\) for some \(\psi: C_9 \to \mathop{\mathrm{Aut}}(C_7)\), and we can take the presentation \begin{align*} G = \left\langle{x, y{~\mathrel{\Big\vert}~}x^7, y^9, yxy^{-1}= \psi(x)}\right\rangle .\end{align*}

• It now suffices to find a nontrivial \(\psi: C_7\to C_7\). Writing it multiplicatively as \(C_7 = \left\langle{x{~\mathrel{\Big\vert}~}x^7}\right\rangle\), any map that sends \(x\) to a generator will do. It suffices to choose any \(k\) coprime to \(7\), and then take \(\psi(x) \coloneqq x^k\), which will be another generator.

• So take

\begin{align*} G = \left\langle{x, y{~\mathrel{\Big\vert}~}x^7, y^9, yxy^{-1}= x^2}\right\rangle .\end{align*}

• We’ll use rank-nullity throughout: \(\operatorname{rank}M + \dim \ker M = 7\).

• Also note that \begin{align*} Av = \lambda v \implies A^nv = A^{n-1}Av = A^{n-1}\lambda v = \cdots = \lambda^n v ,\end{align*} so if \(\lambda \in \operatorname{Spec}(A)\) then \(\lambda^n\in \operatorname{Spec}(A^n)\). Conversely, \(\lambda\in \operatorname{Spec}(A^n) \implies \lambda^{1\over n}\in \operatorname{Spec}(A)\), which we’ll use several times.

• Since \(5 = \operatorname{rank}A = \operatorname{rank}(A - 0\cdot I)\), we have \(\dim \ker(A-0\cdot I) = 2\) contributing an eigenvalue of \(\lambda = 0\) with multiplicity \(2\).

• Since \(4 = \operatorname{rank}A^2 = \operatorname{rank}(A^2 - 0\cdot \lambda) = \operatorname{rank}(A-0\cdot \lambda)^2\), we have that \(\dim \ker(A-0\cdot I)^2 = 3\). Since \(\dim \ker (A-0\cdot I)^1 = 2 < 3\), this means there is 1 generalized eigenvector associated to \(\lambda = 0\).

• Since \(6 = \operatorname{rank}(A-1\cdot I)\), \(\dim \ker (A- 1\cdot I) = 1\), contributing \(\lambda = 1\) with multiplicity 1.

• Since \(\operatorname{rank}(A^3-1\cdot I) = 4\), we have \(\dim \ker (A^3-1\cdot I) = 3\), contributing \(\lambda = 1\) now to \(\operatorname{Spec}(A^3)\) instead of \(\operatorname{Spec}(A)\). Thus some unknown cube roots of 1 are contributed to \(\operatorname{Spec}(A)\), so any of \(1=\zeta_3^0, \zeta_3, \zeta_3^2\) are possibilities at this point. Call these three contributions \(z_1, z_2, z_3\), which may not be distinct.

• Now use that \({\mathrm{tr}}(A) = \sum_{i=1}^n \lambda_i\) is the sum of the diagonal on \(\operatorname{JCF}(A)\), using that trace is a similarity invariant, to write \begin{align*} 1 = {\mathrm{tr}}(A) = (0 + 0) + (0) + (1) + (z_1 + z_2 + z_3) \implies z_1 + z_2 + z_3 = 0 ,\end{align*} which is actually enough to force \(z_1 = 1, z_2 = \zeta_3, z_3 = \zeta_3^2\), since no other combination sums to zero. That \(1 + \zeta_3 + \zeta_3^2 = 0\) is a general fact.

• Since \(\lambda=1\) occurs twice as an eigenvalue but \(\dim \ker(A-1\cdot I) = 1\), the two copies of \(\lambda = 1\) must occur in a nontrivial Jordan block.

• So we get a Jordan form \begin{align*} \operatorname{JCF}(A) = \begin{bmatrix} 0 & & & & & & \\ & 0 & 1 & & & & \\ & & 0 & & & & \\ & & & 1 & 1 & & \\ & & & & 1 & & \\ & & & & & \zeta_3 & \\ & & & & & & \zeta_3^2 \\ \end{bmatrix} .\end{align*}