# Classify the groups of order 1805

Posted on 2010-06-26 by nbloomf

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 $\left\langle \mu, \eta, \omega \;\middle\vert\; \begin{matrix} \mu^{19} = \eta^{19} = \omega^5 = 1, \mu \eta = \eta \mu, \\ \omega \mu = \eta \omega, \omega \eta = \mu^{-1} \eta^4 \omega, \mu \omega = \omega \mu^{-1} \eta^4 \end{matrix} \right\rangle.$