Problem 5.5.9 (Groups of order $1805$)

Proof. Put

We check $A^5=I$:

sage: G = GL(2, GF(19))
sage: G.order()
123120
sage: A = matrix([[0,-1],[1,4]]); A = G(A); A

[ 0 18]
[ 1  4]
sage: A^5

[1 0]
[0 1]

Note that $1805=19^2\cdot 5$.

Consider the map

As in Exercise 8, since $A^5=I$, ${\phi }_{\bar{x}}$ is well defined, and $\phi$ is a homomorphism. Therefore

is a non-abelian group of order $1805$.

The law of $G_0$ is given by

where $a,b,a^{\prime },b^{\prime }\in \mathbb{Z}∕19\mathbb{Z},x,x^{\prime }\in \mathbb{Z}∕5\mathbb{Z}$.

Presentation of $G_0$.

Consider the three elements of $G_0$ defined by

If $\left(\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right),\bar{x}\right)$ is any element of $G_0$, then

so $G_0=⟨a,b,c⟩.$

By (2), we have

Now we prove that

To this end, we must prove that if $\mathrm{\Gamma }$ is any group containing elements $\alpha ,\beta ,\gamma$ such that

then there is a unique homomorphism $\psi :G_0\to \mathrm{\Gamma }$ such that

• Unicity. If $\psi :G_0\to \mathrm{\Gamma }$ is a homomorphism that satisfies (3) and (4), then

  $$\begin{array}{llll}\hfill \psi \left(\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right),\bar{x}\right)& =\psi (a^u{b}^v{c}^w)& \hfill & \\ \hfill & =\psi {(a)}^u\psi {(b)}^v\psi {(c)}^w& \hfill & \\ \hfill & ={\alpha }^u{\beta }^v{\gamma }^w.& \hfill & \end{array}$$

  This proves the unicity of $\psi$.

• Existence. We define the map $\psi :G_0\to \mathrm{\Gamma }$ by

  $$\psi \left(\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right),\bar{x}\right)={\alpha }^u{\beta }^v{\gamma }^w.$$

  Since ${\alpha }^{19}={\beta }^{19}=1$ and ${\gamma }^5=1$ by (3), $\psi$ is a well defined application such that (4) is true.

  We show that $\psi$ is a homomorphism. We prove first by induction on $x$ that for all integers $s,t,w,z$,

  $$\begin{array}{lll}\hfill {\gamma }^x({\alpha }^s{\beta }^t){\gamma }^{-x}={\alpha }^w{\beta }^z\text{where}\left(\begin{array}{c}\hfill \bar{w}\hfill \\ \hfill \bar{z}\hfill \end{array}\right)=A^x\left(\begin{array}{c}\hfill \bar{s}\hfill \\ \hfill \bar{t}\hfill \end{array}\right).& & \hfill \text{(5)}\end{array}$$

  It is true when $x=0$. Suppose that (5) is true for some integer $x\ge 0$. Then

  $$\begin{array}{llll}\hfill {\gamma }^{x+1}({\alpha }^u{\beta }^v){\gamma }^{-(x+1)}& =\gamma ({\gamma }^x{\alpha }^u{\beta }^v{\gamma }^{-x}){\gamma }^{-1}& \hfill & \\ \hfill & =\gamma {\alpha }^w{\beta }^z{\gamma }^{-1}& \hfill & \\ \hfill & =(\gamma {\alpha }^w{\gamma }^{-1})(\gamma {\beta }^w{\gamma }^{-1})& \hfill & \\ \hfill & ={(\mathit{𝛾𝛼}{\gamma }^{-1})}^w{(\mathit{𝛾𝛽}{\gamma }^{-1})}^z& \hfill & \\ \hfill & ={\beta }^w{({\alpha }^{-1}{\beta }^4)}^z& \hfill & \\ \hfill & ={\alpha }^{-z}{\beta }^{w+4z}& \hfill & \\ \hfill & ={\alpha }^{w^{\prime }}{\beta }^{z^{\prime }},& \hfill & \end{array}$$

  where

  $$\left(\begin{array}{c}\hfill {\bar{w}}^{\prime }\hfill \\ \hfill {\bar{z}}^{\prime }\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -\bar{z}\hfill \\ \hfill \bar{w}+4\bar{z}\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 4\hfill \end{array}\right)\left(\begin{array}{c}\hfill \bar{w}\hfill \\ \hfill \bar{z}\hfill \end{array}\right)=A\cdot A^x\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right)=A^{x+1}\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right).$$

  The induction is done, which proves that (5) is true for any $x\ge 0$.

  If $x<0$, by (5) applied to $-x>0$, then ${\gamma }^{-x}{\alpha }^w{\beta }^z{\gamma }^x={\alpha }^s{\beta }^t$, where $\left(\begin{array}{c}\hfill \bar{s}\hfill \\ \hfill \bar{t}\hfill \end{array}\right)=A^{-x}\left(\begin{array}{c}\hfill \bar{w}\hfill \\ \hfill \bar{z}\hfill \end{array}\right)$, thus ${\gamma }^x{\alpha }^s{\beta }^t{\gamma }^{-x}={\alpha }^w{\beta }^z$, where $\left(\begin{array}{c}\hfill \bar{w}\hfill \\ \hfill \bar{z}\hfill \end{array}\right)=A^x\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right)$, so (5) is true for all integers $x$.

  Then, for all $\bar{u},\bar{v}\bar{s},\bar{t}\in \mathbb{Z}∕19\mathbb{Z}$ and for all $\bar{x},\bar{y}\in \mathbb{Z}∕5\mathbb{Z}$,

  $$\begin{array}{llllll}\hfill \psi \left(\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right),\bar{x}\right)\psi \left(\left(\begin{array}{c}\hfill \bar{s}\hfill \\ \hfill \bar{t}\hfill \end{array}\right),\bar{y}\right)& ={\alpha }^u{\beta }^v{\gamma }^x{\alpha }^s{\beta }^t{\gamma }^y& \hfill & & \hfill & \\ \hfill & ={\alpha }^u{\beta }^v[{\gamma }^x({\alpha }^s{\beta }^t){\gamma }^{-x}]{\gamma }^x{\gamma }^y& \hfill & & \hfill & \\ \hfill & ={\alpha }^u{\beta }^v{\alpha }^w{\beta }^z{\gamma }^x{\gamma }^y& \hfill (\text{where }\left(\begin{array}{c}\hfill \bar{w}\hfill \\ \hfill \bar{z}\hfill \end{array}\right)=A^x\left(\begin{array}{c}\hfill \bar{s}\hfill \\ \hfill \bar{t}\hfill \end{array}\right))& & \hfill & & \hfill \\ \hfill & ={\alpha }^{u+w}{\beta }^{v+z}{\gamma }^{x+y},& \hfill & & \hfill & \end{array}$$

  and

  $$\begin{array}{llll}\hfill \psi \left[\left(\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right),\bar{x}\right)\left(\left(\begin{array}{c}\hfill \bar{s}\hfill \\ \hfill \bar{t}\hfill \end{array}\right),\bar{y}\right)\right]& =\psi \left(\left(\begin{array}{c}\hfill \bar{u}\hfill \\ \hfill \bar{v}\hfill \end{array}\right)+A^x\left(\begin{array}{c}\hfill \bar{s}\hfill \\ \hfill \bar{t}\hfill \end{array}\right),\bar{x}+\bar{y}\right)& \hfill & \\ \hfill & =\psi \left(\left(\begin{array}{c}\hfill \overline{u+w}\hfill \\ \hfill \overline{v+z}\hfill \end{array}\right),\overline{x+y}\right)& \hfill & \\ \hfill & ={\alpha }^{u+w}{\beta }^{v+z}{\gamma }^{x+y}.& \hfill & \end{array}$$

  So $\psi$ is a homomorphism.

Put $G_1=⟨A,B,C\mid A^{19}=B^{19}=1,C^5=1,\mathit{𝐴𝐵}=\mathit{𝐵𝐴},\mathit{𝐶𝐴}{C}^{-1}=B,\mathit{𝐶𝐵}{C}^{-1}=A^{-1}{B}^4⟩$. The preceding reasoning shows that there is a homomorphism $\psi :G_0\to G_1$ such that $\psi (a)=A,\psi (b)=B$ and $\psi (c)=C$. By definition of a presentation, there is an homomorphism $\lambda :G_1\to G_0$ such that $\lambda (A)=a,\lambda (B)=b$ and $\lambda (C)=c$. Then $\lambda \circ \psi$ satisfies $(\lambda \circ \psi )(a)=a,(\lambda \circ \psi )(b)=b,(\lambda \circ \psi )(c)=c$. Since $G_0=⟨a,b,c⟩$, this shows that $\lambda \circ \psi =\mathrm{id}$. Similarly, $\psi \circ \lambda =\mathrm{id}$. So $\psi$ is bijective, and $\psi$ is an isomorphism. This proves, with a slight notation modification,

Classification of groups of order $1805$. Let $G$ be a group of order $1805=19^2\cdot 5$. By Theorem 5, if $G$ is abelian,

Now we assume that $G$ is a non-abelian group.

By Sylow’s Theorem, $n_{19}\mid 5$ and $n_{19}\equiv 1(\mathit{𝑚𝑜𝑑}5)$, thus $n_{19}=1$. Let $H$ be the unique Sylow $19$-subgroup, and let $K$ be a $5$-Sylow subgroup. Similarly, $n_5\equiv 1(\mathit{𝑚𝑜𝑑}5)$ and $n_5\mid 19^2$, thus $n_5=1$ or $n_5=19^2$.

If $n_5=1$, then $H⊴G$ and $K⊴G$, and $G=\mathit{𝐻𝐾}\simeq H\times K$ is abelian, in contradiction with our hypothesis. Therefore

Since $H⊴G$, $\mathit{𝐻𝐾}$ is a subgroup of $G$. Moreover $H\cap K=\{1\}$ (if $a\in H\cap K$, then $|a|$ divides $5$ and $19^2$, thus $a=1$). Therefore $G=H{⋊}_{\psi }K$. Moreover, $H\simeq \mathbb{Z}∕19\mathbb{Z}\times \mathbb{Z}∕19\mathbb{Z}$ or $H\simeq \mathbb{Z}∕361\mathbb{Z}$.

If $H\simeq \mathbb{Z}∕361\mathbb{Z}$, then $|\mathrm{Aut}(H)|=\phi (19^2)=19^2-19=342=2\cdot 3^2\cdot 19$, so $g.c.d.(|K|,|\mathrm{Aut}(H)|)=1$. The homomorphism $\psi :K\to \mathrm{Aut}(H)$ is the trivial homomorphism and $G$ is abelian: contradiction. Therefore

for some homomorphism $\xi :\mathbb{Z}∕5\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕19\mathbb{Z}\times \mathbb{Z}∕19\mathbb{Z})\simeq {\mathrm{GL}}_2({𝔽}_{19})$.

There is some matrix $B\in {\mathrm{GL}}_2({𝔽}_{19})$ such that

Since ${\xi }_{\bar{1}}^5=\mathrm{id}$, then $B^5=I$, and since $\xi$ is not trivial, $B\ne I$.

With some help of Sagemath(see APPENDIX) , we obtain that the set

of cardinality $684$, is the union of two disjoint conjugacy classes ${𝒞}_A$ and ${𝒞}_B$, where

Both classes have cardinality $342$.

Moreover $A_1^2\in {𝒞}_A$, and for $Q=\left(\begin{array}{cc}\hfill 10\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 5\hfill \end{array}\right)$, we obtain

With these results, we can show that $G\simeq G_0$:

• If $B\in {𝒞}_A$, then $B=\mathit{𝑃𝐴}{P}^{-1}$ for some $P\in {\mathrm{GL}}_2({𝔽}_{19})$.

• If $B\in {𝒞}_{A_1}$, then $B=R{A}_1{R}^{-1}$ for some $R\in {\mathrm{GL}}_2({𝔽}_{19})$. Therefore

  $$B^2=R{A}_1^2{R}^{-1}=R(\mathit{𝑄𝐴}{Q}^{-1})R^{-1}=(\mathit{𝑅𝑄})A{(\mathit{𝑅𝑄})}^{-1}=\mathit{𝑃𝐴}{P}^{-1}\text{where }P=\mathit{𝑅𝑄}.$$

In both cases there is some integer $k$ ( $k=1$ or $k=2$) and some $P\in {\mathrm{GL}}_3({𝔽}_2)$ such that

If $\sigma \in \mathrm{Aut}({𝔽}_{19}^2)$ is defined by $\sigma (X)=\mathit{𝑃𝑋}$, then for all $X\in {𝔽}_{19}^2$ (where $X^{\prime }=\mathit{𝑃𝑋}=\sigma (X)$), and all $\bar{x}\in \mathbb{Z}∕5\mathbb{Z}$,

Therefore ${\xi }^k(\bar{x})\circ \sigma =\sigma \circ \phi (\bar{x})$ for all $\bar{x}\in K=\mathbb{Z}∕5\mathbb{Z}$, thus

where $K=\mathbb{Z}∕5\mathbb{Z}$ is cyclic.

By Exercise 6,

There is a unique non-abelian group of order $1805$, up to isomorphism.

So there are $3$ types of groups of order $1805$:

where