Let $D_4$ be the dihedral group which is generated by the 2 elements $\rho , \sigma$ subject to the relations $\rho ^4=e$, $\sigma ^2=e$, and $\sigma \rho \sigma \rho =e$ (e denote identity).

$D_4$ has exactly 8 distinct elements, which my for example be written $e, \rho , \rho ^2 , \rho ^3, \sigma , \sigma \rho , \sigma \rho ^2 , \sigma \rho ^3$.

(A) Use the relations to prove that $ \rho \sigma = \sigma \rho ^3$.


(B) Find the elements of $D_4$ which have order 2.

$\textbf{PLEASE CHECK:}$ $ \rho ^2 , \sigma \rho $.

(C) Find the normal subgroups of $D_4$ which have order 2.

$\textbf{PLEASE CHECK:}$ None?

(D) For each normal subgroup N of $D_4$ which has order 2, is $D_4 / N$ isomorphic to $\textbf{Z} / 2 \textbf{Z}$ x $\textbf{Z} / 2 \textbf{Z}$ or $\textbf{Z} / 4 \textbf{Z} $? Explain your reasoning.

I do not understand how to approach this last part.

  • 1
    $\begingroup$ To obtain (A), start with the last of the three relations. Can you use the first two to "isolate" $\rho\sigma$? For (C): a subgroup of order two has only two elements, one of which is the identity. What can you say about the order of the other element? Clearly it will be difficult to approach (D) without having found a subgroup of order two. $\endgroup$ – Adam Saltz Apr 22 '13 at 4:51

For (A), note that \begin{eqnarray*} \sigma\rho\sigma\rho & = & e\\ \sigma\left(\sigma\rho\sigma\rho\right)\rho^{3} & = & \sigma\rho^{3}\\ \sigma^{2}\rho\sigma\rho^{4} & = & \sigma\rho^{3}\\ e\rho\sigma e & = & \sigma\rho^{3}\\ \rho\sigma & = & \sigma\rho^{3} \end{eqnarray*}

For (B), you're right.

For (C), check the subgroup $\left\langle \rho^2 \right\rangle$, $\left\langle \rho\right\rangle $, $\left\{ 1,\rho^{2},\sigma,\sigma\rho^{2}\right\} $, and $\left\{ 1,\rho^{2},\sigma\rho,\sigma\rho^{3}\right\}$.

For (D), use $N=\left\langle \rho^2 \right\rangle$ as an example since it's the only one with order 2. You can see that $$D_4/N \cong \frac{\mathbb{Z}}{2\mathbb{Z}}\times \frac{\mathbb{Z}}{2\mathbb{Z}}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.