Problem 5.5.8 (Groups of order 75)

Question

Construct a non-abelian group of order $75$. Classify all groups of order $75$ (there are three of them).

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\F}{\mathbb{F}}
\begin{proof} 
Let $A \in \mathrm{GL}_2(\mathbb{F}_5)$ be an element of order $3$, for instance
\begin{align}A = \begin{pmatrix} 0 & -1 \ 1 & -1 \end{pmatrix}.
\end{align}
$$(A^2 = \begin{pmatrix} -1 & 1 \-1 & 0 \end{pmatrix}, \qquad A^3 = I.)$$
Consider the map 
\begin{align}
\varphi
\left\{
\begin{array}{ccl}
\Z/3\Z &	o &\mathrm{Aut}(\Z/5\Z 	imes \Z/5\Z) \simeq \mathrm{GL}_2(\F_5)\
\overline{x} = {[}x{]}_3 & \mapsto &
\varphi_{\overline{x}}
    \left\{
    \begin{array}{ccl}
    \Z/5\Z 	imes \Z/5\Z & 	o & \Z/5\Z 	imes \Z/5\Z\
    \begin{pmatrix} a \ b \end{pmatrix} & \mapsto & A^x \begin{pmatrix} a \ b \end{pmatrix}
    \end{array}
    ight.
\end{array}
ight.
\end{align}
Since $A^3 = I$, $\varphi_(\overline{x})$ is well defined. Moreover, for all $\overline{x}, \overline{y} \in \Z/3\Z$, and $a, b \in \Z/5\Z$,
\begin{align*}
\left( \varphi_{\overline{x}} \circ  \varphi_{\overline{y}}ight)\begin{pmatrix} a \ b \end{pmatrix}  &= A^x A^y \begin{pmatrix} a \ b \end{pmatrix} \
&= \varphi_{\overline{x} + \overline{y}} \begin{pmatrix} a \ b \end{pmatrix} ,
\end{align*}
thus $\varphi_{\overline{x}} \circ  \varphi_{\overline{y}} =  \varphi_{\overline{x} + \overline{y}}$, so $\varphi : \Z/3\Z 	o \mathrm{Aut}(\Z/5\Z 	imes \Z/5\Z)$ is a homomorphism.

Hence the group 
$$G_0 = (\Z/5\Z	imes \Z/5\Z) times_{\varphi} \Z/3\Z$$
is well defined. Since $\varphi$ is not the trivial homomorphism, $G$ is not abelian, so there exists a non abelian group of order $75$.

The law in $G_0$ is given by
$$\left(   \begin{pmatrix} a \ b \end{pmatrix} , x ight)\left(   \begin{pmatrix} a' \ b' \end{pmatrix} , x' ight)
= \left(   \begin{pmatrix} a \ b \end{pmatrix} + \begin{pmatrix} -1 & 1 \-1 & 0 \end{pmatrix}^x \begin{pmatrix} a' \ b' \end{pmatrix} , x+x' ight)
$$
where $a,b, a',b' \in \Z/5\Z, \ x,x' \in \Z/3\Z$.
\bigskip

{\bf Classification of groups of order $75$.}

Let $G$ be a group of order $75 = 3 \cdot 5^2$. If $G$ is abelian, by Theorem 5,
$$G \simeq Z_5 	imes Z_5 	imes Z_3 \qquad 	ext{or} \qquad G \simeq Z_{25} 	imes Z_3.$$
Now we assume that $G$ is not abelian.

By Sylow's Theorem, $n_5 \equiv 1 \pmod 5$ and $n_5 \mid 3$, so
$$n_5 = 1.$$
Moreover, $n_3 \equiv 1 \pmod 3$ and $n_3 \mid 5^2$, thus $n_3  = 1$ or $n_3 = 25$.

 Let $H$ be a Sylow $5$- subgroup and $K$ be a Sylow $3$-subgroup. 
 
 Suppose that $n_3 =1$. Then $H \unlhd G$, $K \unlhd G$ and $H \cap K = \{1\}$. Therefore $G = HK \simeq H 	imes K$ is abelian, in contradiction with our assumption.  this shows
$$n_5 = 1,\qquad n_3 = 25.$$
Since $H \unlhd G$ and $H \cap K = \{1\}$,
$$G \simeq H times_{\zeta} K,$$
where $\zeta : K	o  \mathrm{Aut}(H)$ is the homomorphism associated to conjugation. If $\zeta$ is the trivial homomorphism, then $G$ is abelian, in contradiction with our hypothesis. Since $K = \langle x angle$, where $|x| = 3$, then $\zeta$ maps $x$ on an element of order $3$, so $3 \mid \mathrm{Aut}(H)$.

Moreover, $|H| = 5^2$, thus $H$ is abelian, and $H \simeq \Z/25 \Z$ or $H \simeq \Z/5\Z 	imes \Z/5Z$. Since $3 
mid |\mathrm{Aut}(\Z/25\Z)| = 20$, we conclude
$$H \simeq \Z/5\Z 	imes \Z/5\Z.$$ 
Therefore
$$G \simeq (\Z/5\Z 	imes \Z/5\Z)times_{\psi} \Z/3\Z,$$
where
$$\psi: \Z/3\Z 	o \mathrm{Aut}(\Z/5\Z 	imes \Z/5\Z) \simeq \mathrm{GL}_2(\F_5)$$
is a (non trivial) homomorphism.

Then $\psi$ maps the element $\overline{1} \in \Z/3\Z$ of order $3$ on some element $\psi_x$ of order $3$, so there is some matrix $B \in \mathrm{GL}_2(\F_5)$ of order $3$ such that for all $\overline{x} \in \Z/3\Z$, and for all $(a,b) \in \Z/5\Z 	imes \Z/5\Z$,
$$\psi_{\overline{x}}    \begin{pmatrix} a \ b \end{pmatrix} = B^x \begin{pmatrix} a \ b \end{pmatrix} .$$

We prove that $(\Z/5\Z 	imes \Z/5\Z)times_{\psi} \Z/3 \simeq  (\Z/5\Z	imes \Z/5\Z) times_{\varphi} \Z/3\Z$, where $\varphi$ is defined by (1) and (2).

There are $20$ elements of order $3$ in $\mathrm{GL}_2(\F_5)$, and the class of conjugacy of $A$ is the set of these $20$ elements, so there exists $P \in \mathrm{GL}_2(\F_5)$ such that
$$B = P^{-1} A P.$$
(All elements of order $3$ in $\mathrm{GL}_2(\F_5)$ are similar.)

With sagemaths:
\begin{verbatim}
sage: G = GL(2,GF(5))
sage: G.order()
480
sage: l = []
sage: for A in G:
....:     if A != G(1) and A^3 == G(1):
....:         l.append(A)
....:         
sage: l

[
[0 4]  [4 1]  [3 2]  [0 2]  [3 1]  [3 4]  [0 1]  [4 4]  [4 3]  [0 3]
[1 4], [4 0], [1 1], [2 4], [2 1], [3 1], [4 4], [1 0], [3 0], [3 4],

[4 2]  [1 3]  [2 3]  [2 4]  [1 2]  [2 2]  [1 1]  [3 3]  [1 4]  [2 1]
[2 0], [4 3], [1 2], [2 2], [1 3], [4 2], [2 3], [4 1], [3 3], [3 2]
]
sage: A = l[0]; A

[0 4]
[1 4]
sage: A.conjugacy_class().list()

[
[0 4]  [0 3]  [1 1]  [0 1]  [3 3]  [1 2]  [0 2]  [2 2]  [2 1]  [1 4]
[1 4], [3 4], [2 3], [4 4], [4 1], [1 3], [2 4], [4 2], [3 2], [3 3],

[2 4]  [3 1]  [1 3]  [3 2]  [4 4]  [2 3]  [3 4]  [4 3]  [4 2]  [4 1]
[2 2], [2 1], [4 3], [1 1], [1 0], [1 2], [3 1], [3 0], [2 0], [4 0]
]
\end{verbatim}

Since $B = P^{-1}A P$, for all $\overline{x} \in \Z/3\Z$,
$$\psi(\overline{x}) = \sigma \varphi(\overline{x}) \sigma^{-1},$$
where $\sigma : X \mapsto PX$ is an automorphism of $\Z/5\Z	imes \Z/5\Z$ (see Exercise 7).

Thus $\psi(K) = \sigma \varphi(K) \sigma^{-1}$, and by Exercise 6,
$$(\Z/5\Z 	imes \Z/5\Z)times_{\psi} \Z/3 \simeq  (\Z/5\Z	imes \Z/5\Z) times_{\varphi} \Z/3\Z.$$

So there are three types of groups of order $75$, given by
\begin{align*}
&\Z/5\Z 	imes \Z/5\Z	imes \Z/3\Z,\
&\Z/25\Z 	imes\Z/3\Z,\
&(\Z/5\Z	imes \Z/5\Z) times_{\varphi} \Z/3\Z,
\end{align*}
where
\begin{align*}
\varphi
\left\{
\begin{array}{ccl}
\Z/3\Z &	o &\mathrm{Aut}(\Z/5\Z 	imes \Z/5\Z) \simeq \mathrm{GL}_2(\F_5)\
\overline{x} = {[}x{]}_3 & \mapsto &
\varphi_{\overline{x}}
    \left\{
    \begin{array}{ccl}
    \Z/5\Z 	imes \Z/5\Z & 	o & \Z/5\Z 	imes \Z/5\Z\
    \begin{pmatrix} a \ b \end{pmatrix} & \mapsto & A^x \begin{pmatrix} a \ b \end{pmatrix}
    \end{array}
    ight.
\end{array}
ight.
\end{align*}
\end{proof}

Problem 5.5.8 (Groups of order 75)

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Problem 5.5.8 (Groups of order 75)

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