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

  • #1
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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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Answers and Replies

  • #2
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What is ##Q_8##?
 
  • #3
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What is ##Q_8##?
Sorry. It's the quaternion group.
 
  • #4
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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?
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
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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
mathwonk
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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
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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
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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
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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
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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
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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
mathwonk
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