# Subgroup of order $4$ in $D_8$

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!

• I guess you want $\sigma$, say, of order $4$, otherwise you get indeed the Klein four-group, not $D_{8}$ – Andreas Caranti Oct 5 '18 at 17:05
• 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 $D_8$. – Bungo Oct 5 '18 at 17:11

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 $$\langle \sigma^2,\tau\rangle$$.
• 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. – Mark Oct 5 '18 at 19:46
• @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. – Bungo Oct 5 '18 at 19:55
• 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 $H \cap K = 1$. – Bungo Oct 5 '18 at 19:56
• @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. – Mark Oct 5 '18 at 19:57
I think your mistake is to assume that $$\tau$$ and $$\tau \sigma$$ commute. This is not the case $$\tau \cdot \tau \sigma = \sigma,$$ while $$\tau \sigma \cdot \tau = \sigma^{-1} \tau \tau = \sigma^{-1} \ne \sigma.$$