Symmetries of 6-Gon: Proving D_6 isomorphic to D_3 x Z_2

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In summary: T:D_n\times \mathbb{Z}_2\rightarrow D_{2n}, but T(e,1) did not have order 2 in D_n\times \mathbb{Z}_2. In this case, if we try to map T(e,1) onto H, we will get an element that is not in H, which is called a subgroup by convention. In other words, we would have created a new group out of nothing, and it would not be in D_{2n}.
  • #1
CAF123
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Homework Statement


We were given the following argument showing that ##D_6 \cong D_3 \times \mathbb{Z}_2##.
What we did was take two subgroups H and K of G such that their intersection was trivial and that ##hk = kh## for all h and k. H = {symmetries of 6-gon that preserved a triangle inside} and K = <g3>. By some theorem, we proved this shows that HK is isomorphic to H x K which in turn is isomorphic to ##D_3 \times \mathbb{Z}_2##

The question is: Does a similar argument apply to ##D_n## where n is even ##\geq 6##?

The Attempt at a Solution



I presume the question is asking if we can write ##D_n \cong D_x \times \mathbb{Z}_y## for some x and y?

What I have done so far is the following:
Let G = Dn. Choose H and K such that |H| . |K| = 2n. For the above argument to work, necessarily H and K are subgroups of G. So by Lagrange, the orders of H and K have to divide the order of Dn. This means: $$|D_n| = a|H|, |D_n| = b|K|\,\Rightarrow\,a|H| = b|K|.$$Since H and K are subgroups, they are subsets so H,K contained in G. But since we want the intersection to be trivial (i.e ##H \cap K = \left\{e\right\}##, we have that ##|H| \neq |K|##. I am not sure if this helps me at all and I am unsure of how to proceed.

Many thanks.
 
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  • #2
Any ideas? Or is my question difficult to understand?
 
  • #3
I understand that [itex]D_6[/itex] is the dihedral group of order 6?
But what do you mean with [itex]D_3[/itex]?
 
  • #4
Oh, you're using the notation that the order of [itex]D_n[/itex] is 2n. I get it.
 
  • #5
Maybe we should focus first on whether

[tex]D_{2n}\sim D_n\times \mathbb{Z}_2[/tex]

Can you determine centers of both groups? Thus, can you tell me what [itex]Z(D_{2n})[/itex] and what [itex]Z(D_n\times \mathbb{Z}_2)[/itex]is?
 
  • #6
micromass said:
I understand that [itex]D_6[/itex] is the dihedral group of order 6?
But what do you mean with [itex]D_3[/itex]?
H turned out to have the same number of elements as ##D_3##. So we then said there existed an isomorphism ##H \cong D_3## and this led to the conclusion.
 
  • #7
micromass said:
Maybe we should focus first on whether

[tex]D_{2n}\sim D_n\times \mathbb{Z}_2[/tex]

Can you determine centers of both groups? Thus, can you tell me what [itex]Z(D_{2n})[/itex] and what [itex]Z(D_n\times \mathbb{Z}_2)[/itex]is?

Hi micromass,

We covered centres right at the end of our course and I am not familiar with that Z notation. I am doing some revision and this question came quite early on, just after subgroups, cyclic groups, homormorphisms. Is there a way to do it using this?
 
  • #8
OK, so let's say that [itex]D_{2n}\sim D_n\times \mathbb{Z}_2[/itex].

So then we can write [itex]D_{2n}=HK[/itex] with [itex]H\cap K=\{0\}[/itex] and where H and K are both normal. Furthermore, [itex]H\sim D_n[/itex]and [itex]K\sim \mathbb{Z}_2[/itex].

So if it were true that we could write this, then in particular, we could find a normal subgroup of [itex]D_{2n}[/itex] that has order 2. Can you find me all normal subgroups of order 2 in [itex]D_{2n}[/itex]?? Maybe you can start by listing all elements of order 2, and then see if they generate something normal.
 
  • #9
micromass said:
OK, so let's say that [itex]D_{2n}\sim D_n\times \mathbb{Z}_2[/itex].

So then we can write [itex]D_{2n}=HK[/itex] with [itex]H\cap K=\{0\}[/itex] and where H and K are both normal.

Why do they have to be normal?

Furthermore, [itex]H\sim D_n[/itex]and [itex]K\sim \mathbb{Z}_2[/itex].

So if it were true that we could write this, then in particular, we could find a normal subgroup of [itex]D_{2n}[/itex] that has order 2. Can you find me all normal subgroups of order 2 in [itex]D_{2n}[/itex]?? Maybe you can start by listing all elements of order 2, and then see if they generate something normal.

Thanks for your reply, but we didn't really do much with normal subgroups, so I am struggling even to understand why what you said above is true. We had a sort of introductory lecture on things like centres/normal subgroups/class eqn/conjugacy etc..but didn't go into too much detail. Perhaps the element ##h## would have order 2, but I cannot be sure whether this would hold for all even n
 
  • #10
CAF123 said:
Why do they have to be normal?

Because both [itex]D_n[/itex] (or more precisely: [itex]D_n\times\{0\}[/itex]) and [itex]\mathbb{Z}_2[/itex] (or more precisely: [itex]\{e\}\times \mathbb{Z}_2[/itex]) are normal in [itex]D_n\times \mathbb{Z}_2[/itex]. The subgroups of [itex]D_{2n}[/itex] corresponding to [itex]D_n[/itex] (respectively [itex]\mathbb{Z}_2[/itex]) are H (respectively K). So they need to be normal.

Thanks for your reply, but we didn't really do much with normal subgroups, so I am struggling even to understand why what you said above is true. We had a sort of introductory lecture on things like centres/normal subgroups/class eqn/conjugacy etc..but didn't go into too much detail. Perhaps the element ##h## would have order 2, but I cannot be sure whether this would hold for all even n

But let's say we have the following isomorphism: [itex]T:D_n\times \mathbb{Z}_2\rightarrow D_{2n}[/itex], then (e,1) has order 2 in [itex]D_n\times \mathbb{Z}_2[/itex] and thus so has T(e,1). And T(e,1) exactly generates H.

Since you seem to be struggling with basic concepts, I think you have misinterpreted the question. You seem to have interpreted the question as: is [itex]D_{2n}\cong D_x\times \mathbb{Z}_y[/itex]. I don't think this is the way you should interpret it.

I think the question asks you to generalize the exact proof of [itex]D_6\sim D_3\times \mathbb{Z}_2[/itex] to the context [itex]D_{2n}\sim D_n\times \mathbb{Z}_2[/itex]. And they ask whether that proof is still valid. I don't think they want you to actually show the result in general. They just want you to see if the generalization of the proof is valid.
It might be worth contacting your professor and asking him for more clarificiations on the question.
 

FAQ: Symmetries of 6-Gon: Proving D_6 isomorphic to D_3 x Z_2

1. What is a 6-gon?

A 6-gon, also known as a hexagon, is a polygon with six sides and six angles.

2. What does "symmetries of 6-gon" refer to?

The symmetries of a 6-gon refer to the different ways in which the shape can be rotated, reflected, or translated while still maintaining its original form.

3. What is D6 and how is it related to the symmetries of a 6-gon?

D6 is the dihedral group of order 12, which represents all possible symmetries of a regular hexagon. It consists of rotations and reflections that preserve the shape of the hexagon.

4. What is D3 x Z2 and how is it related to the symmetries of a 6-gon?

D3 x Z2 is the direct product of the dihedral group of order 6 and the cyclic group of order 2. It represents the symmetries of a regular hexagon in terms of rotations, reflections, and translations along with additional symmetries from the cyclic group.

5. How is the isomorphism between D6 and D3 x Z2 proven?

The isomorphism between D6 and D3 x Z2 can be proven using group theory and algebraic techniques such as defining a homomorphism and showing that it is both injective (one-to-one) and surjective (onto). It can also be proven by showing that both groups have the same number of elements and follow the same group operations.

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