Problem 5.4.13 ($D_{8n}
ot \simeq D_{4n} imes Z_2$)

Question

Prove that $D_{8n}$ is not isomorphic to $D_{4n} 	imes Z_2$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $D_{8n} = \langle r,s angle$, where $|r| = 4n$, so $D_{8n}$ contains an element of order $4n$.

Every element $h$ of $D_{4n}$ satisfies $h^{2n} = 1$, and every element $k$ of $Z_2$ satisfies $h^2 = 1$, a fortiori $k^{2n} = 1$, therefore every element $x = (h,k)$ of $D_{4n} 	imes Z_2$ satisfies $x^{2n} = 1$, so $D_{4n} 	imes Z_2$ has no element of order $4n$.

This shows that $D_{8n}$ is not isomorphic to $D_{4n} 	imes Z_2$.
\end{proof}

Problem 5.4.13 ($D_{8n} \not \simeq D_{4n} \times Z_2$)

Answers

Comments

Problem 5.4.13 ($D_{8n} \not \simeq D_{4n} \times Z_2$)

Answers

Comments

Add answer