Problem 5.2.15 (Generators of a subgroup of $Z_8 \times Z_4$)

Proof.

By hypothesis

(a)

Then $$\begin{array}{llll}\hfill G=⟨a⟩\times ⟨b⟩& ⟺⟨x⟩\times ⟨y⟩=⟨a⟩\times ⟨b⟩& \hfill & \\ \hfill & ⟺⟨x⟩=⟨a⟩\text{ and }⟨y⟩=⟨b⟩& \hfill & \\ \hfill & ⟺a=x^k\text{ and }b=y^l\text{ where }k,l\not\equiv 0(\mathit{𝑚𝑜𝑑}2).& \hfill & \end{array}$$

There are $8$ solutions:

$$(x,y),(x,y^3),(x^3,y),(x^3,y^3),(x^5,y),(x^5,y^3),(x^7,y),(x^7,y^3).$$

Note: if we search the pairs $(a,b)$ such that $G\simeq ⟨a⟩\times ⟨b⟩$ is the internal product of $⟨a⟩$ and $⟨b⟩$, we find $2\times 112=224$ solutions.

(b)

Since $y^2\notin ⟨x^{2}y⟩=\{1,x^{2}y,x^4{y}^2,x^6{y}^3\}$, then $⟨x^{2}y⟩\cap ⟨y^2⟩=\{1\}$, therefore $$H=⟨x^{2}y,y^2⟩=⟨x^{2}y⟩\cdot ⟨y^2⟩\simeq Z_4\times Z_2.$$

We prove that there are no elements $a$ and $b$ of $G$ such that $G=⟨a⟩\times ⟨b⟩$ and $H=⟨a^2⟩\times ⟨b^2⟩$.

If $H=⟨x^{2}y⟩\times ⟨y^2⟩=⟨a^2⟩\times ⟨b^2⟩$, then

$$⟨x^{2}y⟩=⟨a^2⟩\text{and}⟨y^2⟩=⟨b^2⟩.$$

We write $a=x^k{y}^l$ and $b=x^m{y}^n$, where $0\le k<8,0\le l<4,0\le m<8,0\le n<4$. Then $⟨x^{2}y⟩=⟨a^2⟩$ implies $x^{2}y={(x^{2k}{y}^{2l})}^{\lambda }=x^{2\mathit{𝑘𝜆}}{y}^{2\mathit{𝑙𝜆}}$, thus $y=y^{2\mathit{𝑙𝜆}}$ for some integer $\lambda$, so $1\equiv 2\mathit{𝑙𝜆}(\mathit{𝑚𝑜𝑑}4)$. But this congruence implies $2\mid 1$, which is false, so $H\ne ⟨a^2⟩\times ⟨b^2⟩$. □