# Showing that Q_8 can't be written as a direct product

1. Oct 29, 2018

### Mr Davis 97

1. The problem statement, all variables and given/known data
Prove that the quaternion group $Q_8$ is not isomorphic to a semi-direct product $H\rtimes_\rho K$ for non-trivial groups $H$ and $K$.

2. Relevant equations

3. The attempt at a solution
My idea is to look at the subgroups of $Q_8$ and to show that the intersection of any two nontrivial subgroups is nontrivial. Is this approach valid? My only problem I think is showing that if $Q_8$ were to be isomorphic to a semi-direct product, then $H$ and $K$ must be subgroups of $Q_8$. Is this true in general?

Last edited: Oct 29, 2018
2. Oct 29, 2018

### Staff: Mentor

What is $Q_8$?

3. Oct 29, 2018

### Mr Davis 97

Sorry. It's the quaternion group.

4. Oct 29, 2018

### Staff: Mentor

If we can write $Q_8 \cong H \rtimes_\rho K$ then sure, $H,K \leq Q_8$ are subgroups, $H\cap K=\{\,1\,\}$ and $H \trianglelefteq Q_8$ is normal, and $\rho \, : \,K \longrightarrow \operatorname{Aut}(H)$ a homomorphism. All these conditions are coded in $Q_8 \cong H \rtimes_\rho K$.

We have therefore automatically (per definitionem) a trivial intersection of the two subgroups. If you can show, that this is not possible for any two subgroups, then you are done. I assume you want to show that $1$ cannot be in the intersection without $-1$ being in it as well?

5. Oct 29, 2018

### Mr Davis 97

Oh okay. So is the fact that $H,K$ must subgroups baked into the definition of semi-direct product? I guess the wording of the problem confused me.

So here is my argument:

Suppose that we know the lattice of subgroups of $Q_8$ (say, from a previous exercise): $$\{1\}, \langle -1 \rangle, \langle i \rangle, \langle j \rangle, \langle k \rangle, Q_8.$$ It's clear that $-1$ is an element of all nontrivial subgroups (since $i^2=j^2=k^2=-1$ and clearly $-1\in \langle -1 \rangle$), and so there can never be a trivial intersection of subgroups, and hence we can not form a semi-direct product.

6. Oct 29, 2018

### mathwonk

as i recall, it is a quotient of a semi direct product by an elment of order 2. but i may be wrong after all these years.

7. Oct 29, 2018

### Staff: Mentor

All proper subgroups are either $\mathbb{Z}_2$ or $\mathbb{Z}_4$. Now the set $\mathbb{Z}_4 \times \mathbb{Z}_4$ has too many elements. So only the set $\mathbb{Z}_2 \times \mathbb{Z}_4$ is possible. Yes, $\langle -1 \rangle=\mathbb{Z}_2 \subseteq \mathbb{Z}_4 =\langle i \rangle$, but we have the product only up to isomorphism. So the left part $\mathbb{Z}_2$ in $\mathbb{Z}_2 \times \mathbb{Z}_4$ doesn't need to be identical to the subgroup of the $\mathbb{Z}_4$ part, only isomorphic, which isn't sufficient. W.l.o.g. we have the pairs
$$P:=\{\,(1,1)\; , \;(1,i)\; , \;(1,-1)\; , \;(1,-i)\; , \;(-1,1)\; , \;(-1,i)\; , \;(-1,-1)\; , \;(-1,i)\,\}$$
and we must show, that these are not isomorphic to $Q_8$ regardless how the multiplication is chosen. You see, I still have $(\mathbb{Z}_2,1) \cap (1,\mathbb{Z}_4) = (1,1) = \{\,1\,\}$ although a copy of $\mathbb{Z}_2$ is a subgroup of $\mathbb{Z}_4$. The point is the copy!

Now $\mathbb{Z}_2 \triangleleft Q_8$ as central subgroup. So we have to assume that $Q_8 \cong \mathbb{Z}_2 \rtimes_\rho \mathbb{Z}_4$ for some homomorphism $\rho\, : \,\mathbb{Z}_4 \longrightarrow \operatorname{Aut}(\mathbb{Z}_2)$ and a group multiplication
$$(\varepsilon\; , \;i^n)\cdot (\varepsilon'\; , \;i^m) = (\varepsilon \cdot \rho(i^n)(\varepsilon')\; , \;i^{n+m})$$
on the set $P = \mathbb{Z}_2 \times \mathbb{Z}_4$, and must show that for neither choice of $\rho$ this can be isomorphic to $Q_8$.

The setup has all ingredients which are needed: $(\mathbb{Z}_2,1)\; , \;(1,\mathbb{Z}_4) \leq Q_8\; , \;(\mathbb{Z}_2,1) \trianglelefteq Q_8 \; , \;(\mathbb{Z}_2,1) \cap (1,\mathbb{Z}_4)=1$. Now why can't we choose $\rho \, : \, \mathbb{Z}_4 \longrightarrow \operatorname{Aut}(\mathbb{Z}_2)$ in such a way, that $Q_8 \cong (\mathbb{Z}_2,1)\; \rtimes_\rho \;(1,\mathbb{Z}_4)$ with the multiplication as defined above?

8. Oct 29, 2018

### Mr Davis 97

Before I answer your question, is my argument wrong? I don't see why what I have done is not sufficient. If every subgroup must have $-1$ as an element, then that means that intersections are always nontrivial, so semi-direct products can never be formed, right? Why isn't just writing this suffcient?

9. Oct 29, 2018

### Staff: Mentor

Assume we have a product $G = H \times K$ and $H \subseteq K$. This is no contradiction, because within the product, the elements of $H$ are represented by all pairs $(H,1)$ and those of $K$ by $(1,K)$. These embeddings make the intersection trivial, although as standalone groups, one is a subgroup. The product space is the Cartesian (set) product, which separates the two parts. Simple example: $\mathbb{Z}_2^2$. It is a group with four elements, although both factors are identical. That's why your argument doesn't apply.

But have a closer look on possible $\rho\,!$

10. Oct 29, 2018

### Mr Davis 97

I'm working on looking at the $\rho$, but in the mean time could answer whether this http://www.mathcounterexamples.net/a-group-that-is-no-a-semi-direct-product/ proof makes sense? It's along the lines of what I was trying to do I think.

11. Oct 29, 2018

### Staff: Mentor

That's the same as what we have here. Which $\rho \, : \, \mathbb{Z}_4 \longrightarrow \operatorname{Aut}(\mathbb{Z}_2)$ are possible? What does this mean in the first place, and then in the second? If you answer these questions, you will see the parallels to what you quoted.

12. Nov 29, 2018