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

In summary, the conversation discusses the problem of proving that the quaternion group Q8 is not isomorphic to a semi-direct product of non-trivial groups H and K. The conversation outlines a potential approach, which involves looking at the subgroups of Q8 and showing that the intersection of any two non-trivial subgroups is also non-trivial. However, it is pointed out that this approach may be insufficient, as it does not take into account the possibility of a copy of one subgroup being embedded in the other. The conversation ends with a suggestion to consider a specific case of the problem, where Q8 is isomorphic to a semi-direct product with a specific choice of homomorphism.
  • #1
Mr Davis 97
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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##.

Homework Equations

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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  • #3
fresh_42 said:
What is ##Q_8##?
Sorry. It's the quaternion group.
 
  • #4
Mr Davis 97 said:

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##.

Homework Equations

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?
 
  • #5
fresh_42 said:
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.
 
  • #6
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
Mr Davis 97 said:
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.
Your argumentation is in principle:

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?
 
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  • #8
fresh_42 said:
Your argumentation is in principle:

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?
 
  • #9
Mr Davis 97 said:
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\,!##
 
  • #10
fresh_42 said:
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.
 
  • #11
Mr Davis 97 said:
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.
 
  • #12
my writeup is on pages 43-49 of these notes, but i see i also left that as a problem. sadly, the link my be broken soon.

http://alpha.math.uga.edu/%7Eroy/843-1.pdf
 

1. What is Q_8?

Q_8, or the quaternion group, is a non-abelian group of order 8. It is a mathematical structure that is used to describe rotations in three-dimensional space.

2. What does it mean for Q_8 to be written as a direct product?

A group G is written as a direct product if it can be expressed as the product of two or more subgroups such that every element in G can be written as a product of elements from those subgroups. In the case of Q_8, this means that it can be written as the product of two or more smaller groups.

3. Why is it important to show that Q_8 can't be written as a direct product?

Showing that Q_8 cannot be written as a direct product is important because it helps us understand the structure and properties of this group. It also has applications in other areas of mathematics, such as group theory and abstract algebra.

4. How can we prove that Q_8 cannot be written as a direct product?

There are several ways to prove that Q_8 cannot be written as a direct product. One approach is to use the fact that Q_8 has a specific subgroup structure, and any direct product of subgroups would not have the same structure. Another approach is to use the properties of the group, such as its non-commutativity and lack of normal subgroups, to show that it cannot be decomposed into smaller groups.

5. Are there any other groups that cannot be written as a direct product?

Yes, there are many other groups that cannot be written as a direct product. Some examples include the symmetric group S_n, the alternating group A_n, and the dihedral group D_n. In general, proving that a group cannot be written as a direct product can be a challenging and interesting problem in group theory.

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