Problem 5.5.13 (Groups of order $4p$, $p>3$ prime)

Proof. Let $G$ be a group of order $4p=2^{2}p$, where $p$ is a prime greater than $3$. Let $H\in \mathit{𝑆𝑦}{l}_p(G)$ and $K\in \mathit{𝑆𝑦}{l}_2(G)$, so that $|H|=p$ and $|K|=4$.

By Sylow’s Theorem, $n_p\mid 4$ and $n_p\equiv 1(\mathit{𝑚𝑜𝑑}p)$, thus $n_p=1$ or $n_p=4$. If $n_p=4$, then $4\equiv 1(\mathit{𝑚𝑜𝑑}p)$, thus $p=3$, which is excluded, so

Therefore $H⊴G$, and so $\mathit{𝐻𝐾}$ is a subgroup of $G$. Moreover $H\cap K=\{1\}$, thus $|\mathit{𝐻𝐾}|=|H||K|∕|H\cap K|=4p=|G|$, and so

where $\psi :K\to \mathrm{Aut}(H)$ affords the action of $K$ on $H$ by conjugation.

Since $|H|=p$ and $|K|=4$, $H\simeq \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$ and $K\simeq \mathbb{Z}∕4\mathbb{Z}$ or $K\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$.

• If $K\simeq \mathbb{Z}∕4\mathbb{Z}$, then

  $$G\simeq \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{\phi }\mathbb{Z}∕4\mathbb{Z}$$

  for some homomorphism $\phi :\mathbb{Z}∕4\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})\simeq {(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }\simeq Z_{p-1}$. Then $\phi (1)$ has order $1,2$ or $4$.

  If $|\phi (1)|=1$, then $\phi$ is the trivial homomorphism and

  $$G\simeq \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}\times \mathbb{Z}∕4\mathbb{Z}\simeq \mathbb{Z}∕4\mathit{𝑝\mathbb{Z}}$$

  is cyclic.

  The polynomial $x^2-1=(x-1)(x+1)\in {𝔽}_p[x]$ has two roots $\pm 1$, so the only element of order $2$ in ${(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }$ is $-1$.

  The elements of order $4$ in ${(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }$ are the roots of the polynomial $x^4-1=(x^2-1)(x^2+1)$ distinct of $\pm 1$, so are the roots of $x^2+1$.

  Since $|{(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }|=p-1$, by Lagrange’s Theorem, if there is an element $\alpha$ of order $4$, then $4\mid p-1$, so $p\equiv 1(\mathit{𝑚𝑜𝑑}4)$.

  Conversely, if $p\equiv 1(\mathit{𝑚𝑜𝑑}4)$, then $-1p={(-1)}^{(p-1)∕2}=1$, thus there is a root of $x^2+1\in {𝔽}_4[x]$, so that there exists an element of order $4$, say $\alpha$. Then ${\alpha }^{-1}$ has also order $4$, and $\alpha \ne {\alpha }^{-1}$, otherwise ${\alpha }^2=1$. Then $x^2+1\in {𝔽}_p[x]$ has two solutions, so there are exactly two elements of order $4$ in ${(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }$, $\alpha \text{ and }{\alpha }^{-1}$.

  • Assume $p\equiv 3(\mathit{𝑚𝑜𝑑}4)$.

    There is no element of order $4$ in ${(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }$ and the only element of order $2$ is $-1$, so there is a unique homomorphism $\phi :\mathbb{Z}∕4\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})$, defined by

    $$\phi \{\begin{array}{ccc}\hfill \mathbb{Z}∕4\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕7\mathbb{Z})\hfill \\ \hfill \bar{x}\hfill & \hfill \mapsto \hfill & {\phi }_{\bar{x}}\{\begin{array}{ccc}\hfill \mathbb{Z}∕7\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕7\mathbb{Z}\hfill \\ \hfill a\hfill & \hfill \mapsto \hfill & {(-1)}^{x}a\hfill \end{array}\hfill \end{array}$$

    and the law on $\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{\phi }\mathbb{Z}∕4\mathbb{Z}$ is given by

    $$(a,\bar{x})(b,\bar{y})=(a+{(-1)}^{x}b,\bar{x}+\bar{y}),$$
    $$(a,b\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}},\bar{x},\bar{y}\in \mathbb{Z}∕4\mathbb{Z}).$$

  • Assume $p\equiv 1(\mathit{𝑚𝑜𝑑}4)$.

    There are two elements of order $4$ in ${(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }$, the elements $\alpha$ and ${\alpha }^{-1}$, and the only element of order $2$ is $-1$. The three non trivial homomorphisms ${\phi }_1,{\phi }_2,{\phi }_3:\mathbb{Z}∕4\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})$ are defined for $\bar{x}\in \mathbb{Z}∕4\mathbb{Z}$ and $a\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$ by

    $$\begin{array}{llll}\hfill {\phi }_1(\bar{x})(a)& ={(-1)}^{x}a,& \hfill & \\ \hfill {\phi }_2(\bar{x})(a)& ={\alpha }^{x}a,& \hfill & \\ \hfill {\phi }_3(\bar{x})(a)& ={\alpha }^{-x}a.& \hfill & \end{array}$$

    If $f:\mathbb{Z}∕4\mathbb{Z}\to \mathbb{Z}∕4\mathbb{Z}$ is defined by $f(\bar{x})=-\bar{x}(\bar{x}\in \mathbb{Z}∕4\mathbb{Z})$, then $f$ is an automorphism, and ${\phi }_3={\phi }_2\circ f$, where $f\in \mathrm{Aut}(\mathbb{Z}∕4\mathbb{Z})$. By the Theorem of precomposition by an automorphism (see Appendix of Exercise 6), we obtain $\mathbb{Z}∕\mathit{𝑝𝑍}{⋊}_{{\phi }_2}(\mathbb{Z}∕4Z)\simeq \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_3}\mathbb{Z}∕4\mathbb{Z}$. It remains two semi-direct products $G_1^{\prime }=\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_1}\mathbb{Z}∕4\mathbb{Z}$ and $G_2^{\prime }=\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_2}\mathbb{Z}∕4\mathbb{Z}$, whose laws are defined by

    $$\begin{array}{llllll}\hfill & G_1^{\prime }:(a,\bar{x})(b,\bar{y})=(a+{(-1)}^{x}b,\bar{x}+\bar{y}),& \hfill & & \hfill & \\ \hfill & G_2^{\prime }:(a,\bar{x})(b,\bar{y})=(a+{\alpha }^{x}b,\bar{x}+\bar{y})& \hfill (a,b\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}},\bar{x},\bar{y}\in \mathbb{Z}∕4\mathbb{Z}),& & \hfill & & \hfill \\ \hfill & & \hfill & & \hfill \end{array}$$

    We show that these two groups are not isomorphic.

    In $G_1^{\prime }$, ${(a,\bar{x})}^2=(a(1+{(-1)}^x),2\bar{x})$, so ${(a,\bar{x})}^2=(0,0)$ if and only if $a{(1+(-1))}^x)=0$ and $2\bar{x}=0$, if and only if $2a=0$ and $2\bar{x}=0$. Therefore the only element of order $2$ in $G_1^{\prime }$ is $(0,\overline{\frac{p-1}{2}})$.

    In $G_2^{\prime }$, ${(a,\bar{x})}^2=(a(1+{\alpha }^x),2\bar{x})$, so ${(a,\bar{x})}^2=(0,0)$ if and only if $\bar{x}=0$ and $a=0$, or $\bar{x}=2$ and $a\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$. Therefore there are $p$ elements of order $2$, the elements $(a,2)$ where $a\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$. This shows that

    $$G_1^{\prime }\not\simeq G_2^{\prime }.$$

• If $K\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$, then

  $$G\simeq \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{\phi }(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$$

  for some homomorphism $\phi :\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})\simeq {(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})}^{\times }\simeq Z_{p-1}$.

  Then $\phi (1,0)$ and $\phi (0,1)$ have order $1$ or $2$ in $\mathrm{Aut}(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})$, so $\phi (1,0)\in \{\mathrm{id},\mathrm{neg}\}$, where

  $$\mathrm{neg}\{\begin{array}{ccc}\hfill \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}\hfill & \hfill \to \hfill & \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}\hfill \\ \hfill a\hfill & \hfill \mapsto \hfill & -a\hfill \end{array}$$

  is the unique automorphism of $Z∕\mathit{𝑝\mathbb{Z}}$ of order $2$. Similarly $\phi (0,1)\in \{\mathrm{id},\mathrm{neg}\}$.

  Therefore there are four homomorphisms ${\phi }_i:Z∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\to \mathrm{Aut}(\mathbb{Z}∕\mathit{𝑝\mathbb{Z}})(i=1,2,3,4)$, characterized by

  $$\begin{array}{llll}\hfill {\phi }_1(1,0)=\mathrm{id},& {\phi }_1(0,1)=\mathrm{id},& \hfill & \\ \hfill {\phi }_2(1,0)=\mathrm{neg},& {\phi }_2(0,1)=\mathrm{id},& \hfill & \\ \hfill {\phi }_3(1,0)=\mathrm{id},& {\phi }_3(0,1)=\mathrm{neg},& \hfill & \\ \hfill {\phi }_4(1,0)=\mathrm{neg},& {\phi }_4(0,1)=\mathrm{neg},& \hfill & \end{array}$$

  More explicitly, if $(\bar{a},\bar{b})\in \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ and $x\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$, then

  $$\begin{array}{llll}\hfill {\phi }_1(\bar{a},\bar{b})(x)& =x,& \hfill & \\ \hfill {\phi }_2(\bar{a},\bar{b})(x)& ={(-1)}^{a}x,& \hfill & \\ \hfill {\phi }_3(\bar{a},\bar{b})(\bar{x})& ={(-1)}^{b}x,& \hfill & \\ \hfill {\phi }_4(\bar{a},\bar{b})(\bar{x})& ={(-1)}^{a+b}x.& \hfill & \\ \hfill & & \hfill \end{array}$$

  ${\phi }_1$ is the trivial homomorphism, which gives

  $$G\simeq G_1=\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}\times \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\simeq \mathbb{Z}∕2\mathit{𝑝\mathbb{Z}}\times \mathbb{Z}∕2\mathbb{Z}.$$

  The three others define non-abelian groups

  $$G_2=\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_2}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}),G_3=\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_3}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}),G_4=\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_4}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$$

  which are non isomorphic to $G_1$, since $G_1$ is abelian.

  We show that $G_2,G_3,G_4$ are isomorphic.

  Consider the map

  $$f\{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill \\ \hfill (\bar{a},\bar{b})\hfill & \hfill \mapsto \hfill & (\bar{b},\bar{a})\hfill \end{array}$$

  Then $f$ is an automorphism of $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ (corresponding to the matrix $\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\in {\mathrm{GL}}_2({𝔽}_2)$).

  For all $x\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$, and for all $(\bar{a},\bar{b})\in \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$,

  $$\begin{array}{llll}\hfill ({\phi }_2\circ f)(\bar{a},\bar{b})(x)& ={\phi }_2(\bar{b},\bar{a})(\bar{x})& \hfill & \\ \hfill & ={(-1)}^{b}x& \hfill & \\ \hfill & ={\phi }_3(\bar{a},\bar{b})(x).& \hfill & \end{array}$$

  Therefore ${\phi }_3={\phi }_2\circ f$, where $f\in \mathrm{Aut}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$. By the Theorem of precomposition by an automorphism (see Appendix of Exercise 6), we obtain $\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_2}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\simeq \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_3}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$, and so

  $$G_2\simeq G_3.$$

  Now consider

  $$g\{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill \\ \hfill (\bar{a},\bar{b})\hfill & \hfill \mapsto \hfill & (\bar{a}+\bar{b},\bar{b})\hfill \end{array}$$

  Then $g$ is an automorphism of $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ (corresponding to the matrix $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in {\mathrm{GL}}_2({𝔽}_2)$). For all $x\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}$, and for all $(\bar{a},\bar{b})\in \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$,

  $$\begin{array}{llll}\hfill ({\phi }_2\circ g)(\bar{a},\bar{b})(x)& ={\phi }_2(\bar{a}+\bar{b},\bar{b})(x)& \hfill & \\ \hfill & ={(-1)}^{a+b}x& \hfill & \\ \hfill & ={\phi }_4(\bar{a},\bar{b})(x).& \hfill & \end{array}$$

  Therefore ${\phi }_4={\phi }_2\circ g$, where $g\in \mathrm{Aut}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$. By the Theorem of precomposition by an automorphism (see Appendix of Exercise 6), we obtain $\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_2}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})\simeq \mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_4}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z})$, and so

  $$G_2\simeq G_4.$$

In conclusion, every group of order $4p$ (where $p$ is a prime and $p>3$) is isomorphic to one of the following groups:

If $p\equiv 3(\mathit{𝑚𝑜𝑑}4)$,

• $Z_{4p}$,
• $Z_{2p}\times Z_2$,

• $\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_1}\mathbb{Z}∕4\mathbb{Z},$ with the law

  $$(a,\bar{x})(b,\bar{y})=(a+{(-1)}^{x}b,\bar{x}+\bar{y})(a,b\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}},\bar{x},\bar{y}\in \mathbb{Z}∕4\mathbb{Z}).$$

• $Z∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_2}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}),$ with the law

  $$(x,\bar{a},\bar{b})(y,\bar{c},\bar{d})=(x+{(-1)}^{a}y,\bar{a}+\bar{c},\bar{b}+\bar{d})(x,y\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}},\bar{a},\bar{b},\bar{c},\bar{d}\in \mathbb{Z}∕2\mathbb{Z}).$$

if $p\equiv 1(\mathit{𝑚𝑜𝑑}4)$,

• $Z_{4p}$,
• $Z_{2p}\times Z_2$,

• $\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_1}\mathbb{Z}∕4\mathbb{Z},$ with the law

  $$(a,\bar{x})(b,\bar{y})=(a+{(-1)}^{x}b,\bar{x}+\bar{y})(a,b\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}},\bar{x},\bar{y}\in \mathbb{Z}∕4\mathbb{Z}).$$

• $\mathbb{Z}∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_2}\mathbb{Z}∕4\mathbb{Z},$ with the law

  $$(a,\bar{x})(b,\bar{y})=(a+{\alpha }^{x}b,\bar{x}+\bar{y})(a,b\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}},\bar{x},\bar{y}\in \mathbb{Z}∕4\mathbb{Z}).$$

  (where $\alpha$ is any root of $x^2+1\in {𝔽}_p[x]$)

• $Z∕\mathit{𝑝\mathbb{Z}}{⋊}_{{\phi }_2}(\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}),$ with the law

  $$(x,\bar{a},\bar{b})(y,\bar{c},\bar{d})=(x+{(-1)}^{a}y,\bar{a}+\bar{c},\bar{b}+\bar{d})(x,y\in \mathbb{Z}∕\mathit{𝑝\mathbb{Z}},\bar{a},\bar{b},\bar{c},\bar{d}\in \mathbb{Z}∕2\mathbb{Z}).$$