1
$\begingroup$

Let me ask you a question on group theory which confuses me.

Consider the group $D_8$ the dihedral group of order $8$ generated by $\sigma$ and $\tau$ with $o(\sigma)=4,o(\tau)=2$ and $\tau\sigma=\sigma^{-1}\tau$.

Consider the following elements, namely $\sigma\tau$ and $\tau$ which have order equal to $2$. Then $\langle\tau\rangle \cong \mathbb{Z}_2$ and $\langle\sigma\tau\rangle \cong \mathbb{Z}_2$ then $\langle\tau\rangle \times \langle\sigma\tau\rangle \cong \mathbb{Z}_2\times \mathbb{Z}_2$. We know that $\mathbb{Z}_2\times \mathbb{Z}_2$ is Klein group. But I checked that the set $\langle\tau\rangle \times \langle\sigma\tau\rangle$ is not even group because $\tau\sigma\tau=\sigma^{-1}$ which does not belong to this set.

Where is the problem?

Would be very grateful for explanation!

$\endgroup$
  • $\begingroup$ I guess you want $\sigma$, say, of order $4$, otherwise you get indeed the Klein four-group, not $D_{8}$ $\endgroup$ – Andreas Caranti Oct 5 '18 at 17:05
  • $\begingroup$ If you seek a Klein subgroup, consider $\langle \tau \rangle \langle \sigma^2 \rangle$ instead. This is indeed a subgroup because $\langle \sigma^2 \rangle$ is normal in $D_8$. $\endgroup$ – Bungo Oct 5 '18 at 17:11
2
$\begingroup$

You are right, $\langle\tau\rangle\times \langle\sigma\tau\rangle\cong\mathbb{Z_2}\times\mathbb{Z_2}$. But it is just not a subgroup of the dihedral group. The elements of $\langle\tau\rangle\times \langle\sigma\tau\rangle$ are ordered pairs of elements of $D_4$, so it is a subgroup of $D_4\times D_4$. Though $D_4$ really has a subgroup which is isomorphic to $\mathbb{Z_2}\times\mathbb{Z_2}$, check $\langle \sigma^2,\tau\rangle$.

$\endgroup$
  • $\begingroup$ Thanks for reply! But some things are still unclear to me. If its isomorphic to Klein group then it is group. But as wee this is not group since not closed. Could you explain more in detais? I would be very grateful for that! $\endgroup$ – ZFR Oct 5 '18 at 19:34
  • $\begingroup$ Yes, $\langle\tau\rangle\times\langle\sigma\tau\rangle$ is a group with respect to operation in every coordinate. Direct product of groups is always a group. Maybe you are confusing between $\langle\tau\rangle\times\langle\sigma\tau\rangle$ and $\langle\tau\rangle\langle\sigma\tau\rangle$? Well, $\langle\tau\rangle\langle\sigma\tau\rangle$ is really not a group and hence it is of course not isomorphic to Klein group. $\endgroup$ – Mark Oct 5 '18 at 19:46
  • $\begingroup$ @RFZ (Sorry, fixing an error in my previous comment.) $\langle \tau \rangle \times \langle \sigma \tau \rangle$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$. But $\langle \tau \rangle \times \langle \sigma \tau \rangle$ is an external direct product. $D_4$ doesn't contain $\langle \tau \rangle \times \langle \sigma \tau \rangle$. It does contain the product $\langle \tau \rangle \langle \sigma \tau \rangle$. But this product isn't even a subgroup, let alone a direct product. $\endgroup$ – Bungo Oct 5 '18 at 19:55
  • $\begingroup$ In general, if $G$ is a group and $H$ and $K$ are subgroups, then you can always form the product $HK$ as a subset of $G$, but this isn't guaranteed to be a subgroup unless at least one of $H$ or $K$ is normal. (In your case, neither $\langle \tau \rangle$ nor $\langle \sigma \tau \rangle$ is normal.) And even if $HK$ is a subgroup, it is not isomorphic to $H \times K$ unless $H$ and $K$ are both normal and $H \cap K = 1$. $\endgroup$ – Bungo Oct 5 '18 at 19:56
  • $\begingroup$ @Bungo, one of the groups being normal is not necessary though. $HK$ is a subgroup if and only if $HK=KH$. One of $H$ and $K$ being normal in $G$ just makes it happen for sure. $\endgroup$ – Mark Oct 5 '18 at 19:57
1
$\begingroup$

I think your mistake is to assume that $\tau$ and $\tau \sigma$ commute. This is not the case $$ \tau \cdot \tau \sigma = \sigma, $$ while $$ \tau \sigma \cdot \tau = \sigma^{-1} \tau \tau = \sigma^{-1} \ne \sigma. $$

$\endgroup$

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.