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

## Homework Statement

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 $K$.

## 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?

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fresh_42
Mentor
What is $Q_8$?

What is $Q_8$?
Sorry. It's the quaternion group.

fresh_42
Mentor

## Homework Statement

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 $K$.

## 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?
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?

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?
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.

mathwonk
Homework Helper
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.

fresh_42
Mentor
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.

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?

• Mr Davis 97

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?
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?

fresh_42
Mentor
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?
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 $\rho\,!$

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 $\rho\,!$
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.

fresh_42
Mentor
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.
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.

mathwonk