Exercise 2.3.13 ($\mathbb{Z} imes Z_2
ot \simeq \mathbb{Z}$ and $\mathbb{Q} imes Z_2
ot \simeq \mathbb{Q}$ )

Question

Prove that the following pair of groups are {\bf not} isomorphic:
\begin{enumerate}
\item[{\bf (a)}] $\mathbb{Z} 	imes Z_2$ and $\mathbb{Z}$
\item[{\bf (b)}] $\mathbb{Q} 	imes Z_2$ and $\Q$.
\end{enumerate}

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

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

\begin{proof} 
\begin{enumerate}
\item[{\bf (a)}] The group $\mathbb{Z} 	imes Z_2$ is not cyclic by Exercise 13 (where we replace  $Z_2 	imes \mathbb{Z}$ by the isomorphic group $\mathbb{Z} 	imes Z_2$), but $\mathbb{Z}$ is cyclic. If a group $G$ is cyclic and $G' \simeq G$, then $G'$ is cyclic. Therefore
$$\mathbb{Z} 	imes Z_2
ot \simeq \mathbb{Z}.$$
\item[{\bf (b)}] The order of every element of $\Q$ is $1$ or $\infty$. But the order of $(0,\overline{1}) \in \mathbb{Q} 	imes Z_2$ is $2$. Hence
$$\mathbb{Q} 	imes Z_2
ot \simeq \Q.$$

\end{enumerate}

\end{proof}

Exercise 2.3.13 ($\mathbb{Z} \times Z_2\not \simeq \mathbb{Z}$ and $\mathbb{Q} \times Z_2 \not \simeq \mathbb{Q}$ )

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Exercise 2.3.13 ($\mathbb{Z} \times Z_2\not \simeq \mathbb{Z}$ and $\mathbb{Q} \times Z_2 \not \simeq \mathbb{Q}$ )

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