0
$\begingroup$

We have the following subgroup of the dihedral group $D_6=\langle \sigma , \tau \mid \sigma^6=id=\tau^2, \tau\sigma=\sigma^5\tau\rangle$ $$\{\langle \sigma^2\rangle, \langle \sigma^2\rangle\sigma , \langle \sigma^2\rangle\tau , \langle \sigma^2\rangle\sigma \tau\}$$

This subgroup has $4$ elements, so it is isomorphic either to $\mathbb{Z}_4$ or to $\mathbb{Z}_2\times\mathbb{Z}_2$. To check to which of them it is isomorphic to, we have to check the order of each element, or not? Do all the elements of the above subgroup have the same order?

Also, an other subgroup is $$\{\langle \sigma^3\rangle, \langle \sigma^3\rangle\sigma , \langle \sigma^3\rangle\sigma^2, \langle \sigma^3\rangle\tau, \langle \sigma^3\rangle\sigma\tau, \langle \sigma^3\rangle\sigma^2\tau\}$$

This subgroup has $6$ elements, so it is isomorphic either to $\mathbb{Z}_6$ or to $D_3$. To check to which of them it is isomorphic to, what do we have to do?

$\endgroup$
  • 1
    $\begingroup$ I don't really understand your notation but notice $\mathbb{Z}_4$ has an order $4$ element while the other group of size $4$ doesn't. Similarly, $\mathbb{Z}_6$ has an order $6$ element while $D_3$ does not. $\endgroup$ – ET93 Nov 25 '16 at 3:47
  • $\begingroup$ Ah ok... And how can we compute the order of each element in this case? @ET93 $\endgroup$ – Mary Star Nov 25 '16 at 3:51
  • 1
    $\begingroup$ Those aren't subgroups but quotient groups. You compute order as you usually would but as soon as something is a power of $\sigma^2$, its the identity in the quotient. $\endgroup$ – ET93 Nov 25 '16 at 3:55
1
$\begingroup$

For the first quotient,

$(\langle \sigma^2 \rangle \sigma)^2=\langle \sigma^2 \rangle \sigma^2=\langle \sigma^2 \rangle$ since $\sigma^2 \in \langle \sigma^2 \rangle$.

$(\langle \sigma^2 \rangle \tau)^2=\langle \sigma^2 \rangle \tau^2=\langle \sigma^2 \rangle \,id=\langle \sigma^2 \rangle$

$(\langle \sigma^2 \rangle \sigma \tau)^2=\langle \sigma^2 \rangle \sigma \tau \sigma \tau=\langle \sigma^2 \rangle \sigma \sigma^5 =\langle \sigma^2 \rangle \,id=\langle \sigma^2 \rangle$

So everything has order $2$ and so this quotient is $\mathbb{Z}_2 \times \mathbb{Z}_2$. The second quotient in a similar way is $D_3$.

$\endgroup$
  • $\begingroup$ I understand!! Thank you very much!! :-) $\endgroup$ – Mary Star Nov 25 '16 at 4:12

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.