Classify the groups of order 1805

Posted on 2010-06-26 by nbloomf
Tags: classification, finite-group, group-presentation, semidirect-product, d-and-f

Classify groups of order \(1805 = 5 \cdot 19^2\).

By FTFGAG, the abelian groups of order 1805 are (up to isomorphism) \(Z_{1805}\) and \(Z_{95} \times Z_{19}\).

Now suppose \(G\) is a nonabelian group of order 1805. By Sylow’s Theorem, \(n_{19} = 1\); let \(H\) denote the unique (hence normal) Sylow 19-subgroup of \(G\). Note that \(H\) must be abelian. Now let \(K = \langle x \rangle \cong Z_5\) be a Sylow 5-subgroup of \(G\). By Lagrange we have \(H \cap K = 1\), so that \(HK = G\). By the recognition theorem for semidirect products we have \(G = H \rtimes_\psi K\) for some \(\psi : K \rightarrow \mathsf{Aut}(H)\). Note that \(\psi\) must be nontrivial, as otherwise \(G \cong H \times K\) is abelian. There are two groups of order \(19^2\) up to isomorphism, and thus two possibilities for the isomorphism type of \(H\): \(Z_{361}\) and \(Z_{19}^2\).

Suppose first that \(H \cong Z_{361}\). We have that \(\mathsf{Aut}(Z_{361}) \cong Z_{19 \cdot 18}\), which has no element of order 5 by Lagrange. In this case every group homomorphism \(\psi : K \rightarrow \mathsf{Aut}(H)\) is trivial, a contradiction.

Thus we may assume that \(H \cong Z_{19}^2\). Now \(\mathsf{Aut}(H) \cong \mathsf{GL}_2(\mathbb{F}_{19})\), and this group has order \(123120 = 2^4 \cdot 3^4 \cdot 5 \cdot 19\). In particular, the Sylow 5-subgroups of \(\mathsf{Aut}(H)\) have order 5. Because \(\psi\) is nontrivial and \(Z_5\) is simple, \(\mathsf{im}\ \psi\) is a Sylow 5-subgroup of \(\mathsf{Aut}(H)\). By Sylow’s Theorem, all Sylow 5-subgroups are conjugate in \(\mathsf{Aut}(H)\). Since \(Z_5\) is cyclic, by a previous exercise, the choice of \(\psi\) does not affect the isomorphism type of \(G\). Thus we have \(G \cong Z_{19}^2 \rtimes_\psi Z_5\), where \(\psi\) is any injective homomorphism \(Z_5 \rightarrow \mathsf{Aut}(Z_{19}^2)\).

To summarize, there are three groups of order 1805 as follows.

  1. \(Z_{1805}\)
  2. \(Z_{95} \times Z_{19}\)
  3. \(Z_{19}^2 \rtimes_\psi Z_5\), where \(\psi : Z_5 \rightarrow \mathsf{Aut}(Z_{19}^2)\) is any injective homomorphism.

Next we will construct an explicit group of the third type and find a presentation for it. First we identify \(Z_{19}^2\) with the additive group of the two-dimensional vector space \(\mathbb{F}_{19}^2 = H\), and identify \(\mathsf{Aut}(Z_{19}^2)\) with \(\mathsf{GL}_2(\mathbb{F}_{19})\). Let \(A\) be the matrix \[A = \begin{bmatrix} 0 & -1 \\ 1 & 4 \end{bmatrix} \in \mathsf{GL}_2(\mathbb{F}_{19}).\] Evidently we have \[\begin{array}{rcl} A^2 & = & \begin{bmatrix} -1 & -4 \\ 4 & -4 \end{bmatrix}, \\ A^3 & = & \begin{bmatrix} -4 & 4 \\ -4 & -1 \end{bmatrix}, \\ A^4 & = & \begin{bmatrix} 4 & 1 \\ -1 & 0 \end{bmatrix},\ \mathrm{and} \\ A^5 & = & I, \end{array}\] so that \(A\) has order 5.

Let \(x\) denote a generator of the cyclic group \(Z_5 = K\). By a prior result, there is a unique group homomorphism \(\varphi : Z_5 \rightarrow \mathsf{GL}_2(\mathbb{F}_{19})\) such that \(\varphi(x) = A\). Define \(G = \mathbb{F}_{19}^2 \rtimes_\varphi Z_5\); certainly this group has order 1805.

Now we find a presentation for this group. First, note that in general if \(H\) is generated by \(S\) and \(K\) is generated by \(T\), then (identifying factors with subgroups) the semidirect product \(H \rtimes K\) is generated by \(S \cup T\). In this case, \(G\) is generated by \[\mu = \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}, 1\right), \eta = \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}, 1\right),\ \mathrm{and}\ \omega = \left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, x\right).\]

Certainly \(\mu \eta = \eta \mu\).

We have \[\omega \mu = (1,x)\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}, 1\right) = \left(A\begin{bmatrix} 1 \\ 0 \end{bmatrix}, x\right) = \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}, x\right) = \eta \omega\] and similarly \[\omega \eta = (1,x)\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}, 1\right) = \left(\begin{bmatrix} -1 \\ 4 \end{bmatrix}, x\right) = \mu^{-1} \eta^4 \omega.\]

Finally, note that \(A \begin{bmatrix} -1 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\); it is straightforward to show that \(\mu \omega = \omega \mu^{-1} \eta^4\).

Thus \(G\) has the presentation \[\left\langle \mu, \eta, \omega \;\middle\vert\; \begin{matrix} \mu^{19} = \eta^{19} = \omega^5 = 1, \mu \eta = \eta \mu, \\ \omega \mu = \eta \omega, \omega \eta = \mu^{-1} \eta^4 \omega, \mu \omega = \omega \mu^{-1} \eta^4 \end{matrix} \right\rangle.\]