Exercise 3.1.21 ($Z_4 \times Z_4/\langle x^2y^2\rangle \simeq Z_4 \times Z_2$)

Proof. Here

so that

We define

(a)

Since $|x^2{y}^2|=2$, then $⟨x^2{y}^2⟩=\{1,x^2{y}^2\}$, thus $$|\bar{G}|=|G|∕|⟨x^2{y}^2⟩|=16∕2=8.$$

(b)

By (1), every element $\bar{z}\in \bar{G}$ is of the form $\bar{z}={\bar{x}}^a{\bar{y}}^b$, where $0\le a<4,0\le b<4$. Since $|\bar{G}|=8$, there are duplicates. Since ${\bar{x}}^2{\bar{y}}^2=1$, then ${\bar{y}}^2={\bar{x}}^{-2}={\bar{x}}^2$. Removing these duplicates (for instance ${\bar{x}}^3{\bar{y}}^3={\bar{x}}^3{\bar{x}}^2\bar{y}=\bar{x}\bar{y}$), it remains $$\bar{G}=\{1,\bar{x},{\bar{x}}^2,{\bar{x}}^3,\bar{y},\bar{x}\bar{y},{\bar{x}}^2\bar{y},{\bar{x}}^3\bar{y}\}.$$

Since $|\bar{G}|=8$, these eight elements are distinct, so every element $\bar{z}\in \bar{G}$ has a unique writing in the form

$$\bar{z}={\bar{x}}^a{\bar{y}}^b,0\le a<4,0\le b<2.$$

(Note that we can exchange the roles of $x$ and $y$ to write every element of $\bar{G}$ in the form $x^a{y}^b,0\le a<2,0\le b<4$.)

(c)

Since ${\bar{x}}^4=\bar{1}$ and ${\bar{x}}^2\ne \bar{1}$ by part (b), $|\bar{x}|=4$, therefore $|{\bar{x}}^i|=\frac{4}{4\wedge i},i=0,1,2,3$.

Note that for all integers $i$, ${({\bar{x}}^i\bar{y})}^4={({\bar{x}}^4)}^i{\bar{y}}^4=1$, and

$$\begin{array}{llll}\hfill {({\bar{x}}^i\bar{y})}^2=1& ⟺{\bar{x}}^{2i}{\bar{y}}^2=1& \hfill & \\ \hfill & ⟺{\bar{x}}^{2i}{x}^2=1& \hfill & \\ \hfill & ⟺{\bar{x}}^2(i+1)=1& \hfill & \\ \hfill & ⟺4\mid 2(i+1)& \hfill & \\ \hfill & ⟺2\mid i+1& \hfill & \\ \hfill & ⟺i\equiv 1(\mathit{mod}2).& \hfill & \end{array}$$

Therefore $\bar{x}\bar{y}$ and ${\bar{x}}^3\bar{y}$ have order $2$, and $\bar{y}$ and ${\bar{x}}^2\bar{y}$ have order $4$. Explicitly,

$x$	$\bar{1}$	$\bar{x}$	$\overline{x^2}$	$\overline{x^3}$	$\bar{y}$	$\overline{\mathit{xy}}$	$\overline{x^{2}y}$	$\overline{x^{3}y}$
$|x|$	$1$	$4$	$2$	$4$	$4$	$2$	$4$	$2$

(d)

The elements $\bar{x}$ and $\overline{\mathit{xy}}$ satisfy $\bar{G}=⟨\bar{x},\overline{\mathit{xy}}⟩$ and $${\bar{x}}^4=\bar{1},{\overline{\mathit{xy}}}^2=\bar{1},\bar{x}(\overline{\mathit{xy}})=(\overline{\mathit{xy}})\bar{x}.$$

Since $Z_4\times Z_2=⟨a,b\mid a^4=b^2=1,\mathit{ab}=\mathit{ba}⟩$, by van Dyck’s Theorem there is a surjective homomorphism $\phi :Z_4\times Z_2\to \bar{G}$ such that $\phi (a)=\bar{x}$ and $\phi (b)=\overline{\mathit{xy}}$. Moreover $|Z_4\times Z_2|=|\bar{G}|=8$, thus $\phi$ is an isomorphism, and

$$\bar{G}\simeq Z_4\times Z_2.$$

Alternatively, without presentations of groups, consider the map

$$\psi \{\begin{array}{ccc}\hfill \mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \bar{G}\hfill \\ \hfill ([i{\text{]}}_4,[j{\text{]}}_2)\hfill & \hfill \mapsto \hfill & {\bar{x}}^i{(\overline{\mathit{xy}})}^j\hfill \end{array}$$

Then

• $\psi$ is well defined: If $i\equiv i^{\prime }(\mathit{mod}4)$ and $j\equiv j^{\prime }(\mathit{mod}2)$, then ${\bar{x}}^i={\bar{x}}^{i^{\prime }}$ and ${\overline{\mathit{xy}}}^j={\overline{\mathit{xy}}}^{j^{\prime }}$, so ${\bar{x}}^i{(\overline{\mathit{xy}})}^j={\bar{x}}^{i^{\prime }}{(\overline{\mathit{xy}})}^{j^{\prime }}$.

• $\psi$ is surjective: By part (b), every element $\bar{z}\in \bar{G}$ is of the form $\bar{z}={\bar{x}}^a{\bar{y}}^b$, where $0\le a<4,0\le b<2$, thus

  $$\bar{z}={\bar{x}}^{a-b}{(\overline{\mathit{xy}})}^b=\psi ({[a-b]}_4,{[b]}_2).$$

• $\psi$ is a homomorphism: if $([i{\text{]}}_4,[j{\text{]}}_2)\in \mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ and $([k{\text{]}}_4,[l{\text{]}}_2)\in \mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$, then

  $$\begin{array}{llll}\hfill \psi (([i{\text{]}}_4,[j{\text{]}}_2)+([k{\text{]}}_4,[l{\text{]}}_2))& =\psi (([i+k{\text{]}}_4,[j+l{\text{]}}_2)& \hfill & \\ \hfill & ={\bar{x}}^{i+k}{(\overline{\mathit{xy}})}^{j+l}& \hfill & \\ \hfill & =({\bar{x}}^i{(\overline{\mathit{xy}})}^j)({\bar{x}}^k{(\overline{\mathit{xy}})}^l)& \hfill & \\ \hfill & =\psi (([i{\text{]}}_4,[j{\text{]}}_2))\psi (([k{\text{]}}_4,[l{\text{]}}_2)).& \hfill & \end{array}$$

• Since $\psi :\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\to \bar{G}$ is a surjective homomorphism, where $|\mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}|=|\bar{G}|=8$, then $\psi$ is an isomorphism, so

  $$\bar{G}\simeq \mathbb{Z}∕4\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}\simeq Z_4\times Z_2.$$