What Are the Subgroups and Homomorphisms of D5 from Z/n for n=1 to 10?

[nhartung]

Homework Statement

Consider the group D₅, the set of all twists and flips which we can perform on a regular pentagonal plate to pass through a fixed regular pentagonal hole under composition.

a. Find all subgroups of D₅ of order 2, if order 3, and of order 5.

b. Find all homomorphisms: $$Z$$$$/$$n $$\rightarrow$$ D₅ for each n from 1 to 10. (That is Z(mod)n not Z divides n, Z being the integers)

Homework Equations

None

The Attempt at a Solution

Ok, I think I have a figured out but I'm completely stuck on b, I'm having trouble determining what a homomorphism is.

For my solution for a I found all of the possible subgroups of D₅ and looked at their order. I came up with this:

For reference:
<T> = Turn once
<T²> = Turn twice
etc.
<FL> = Flip
<FLT> = Flip then Turn
etc.

<T> = {T,T²,T³, T⁴,e}
<T²> = {T², T⁴, T, T³, e}
<T³> = {T³, T, T⁴, T², e}
<T⁴> = {T⁴, T³, T², T, e}

Right away I noticed that these are all creating the same subgroup of order 5.

<FL> = {FL, e}
<FLT> = {FLT, e}
<FLT²> = {FLT², e>
<FLT³> = {FLT³, e>
<FLT⁴> = {FLT⁴, e>

These are all separate subgroups of order 2.

There are no subgroups of order 3.

How do you guys think this looks for part a? Hopefully I'm doing this correctly.

For part b I'm completely lost. I see how D₅ is working like Z(mod)5 and Z(mod)2 I just don't know how find a homomorphism. Maybe if you could give me an example similar to the problem I have here I could make some sense of it.

Thanks

 

[ystael]

Your solution to part a is correct: there is one subgroup of order five generated by the rotation $$\rho$$, and five subgroups of order two generated by each of the five elements $$\sigma, \sigma\rho, \sigma\rho^2, \sigma\rho^3, \sigma\rho^4$$, where $$\sigma$$ is any flip.

For part b, you need some basic facts about how subgroups work. Do you know Lagrange's theorem, that the order of a subgroup divides the order of the group? Use this, and the fact that the kernel of a homomorphism $$\phi: G \to H$$ is a subgroup of $$G$$ and the image of $$\phi$$ is a subgroup of $$H$$. These should help you determine what the homomorphisms $$\mathbb{Z}/n\mathbb{Z} \to D_5$$ are.

 

[nhartung]

I've been looking at what you said for awhile and it doesn't seem to be getting me anywhere but here are some attempts anyway. In class we proved that for any homomorphism f:(G, *) $$\rightarrow$$ (H, *) must take the identity of G to the identity of H.

I know the identity for both of these groups is 0 but I'm not sure how it would work for Z/9 where i choose Z = 9. In this case I get the identity element in Z/9 but in D₅ I get T⁴. Does this mean that Z/9 is not a homomorphism? (And by the same reasoning Z/7, Z/3 and Z/1). Other than this I haven't had much progress.

I'm not sure how I would use Lagrange's theorm here. My teacher didnt do any example of finding homomorphisms in class, he only defined them and I'm no math major so I'm completely lost here.

 

[ystael]

I know the identity for both of these groups is 0 but I'm not sure how it would work for Z/9 where i choose Z = 9. In this case I get the identity element in Z/9 but in D₅ I get T⁴. Does this mean that Z/9 is not a homomorphism? (And by the same reasoning Z/7, Z/3 and Z/1). Other than this I haven't had much progress.

I'm afraid there is a lot of nonsense in here -- you seem to be really confused about some things.

I know the identity for both of these groups is 0

The identity element of $$D_5$$ is the identity map $$\iota$$ (the transformation of the pentagon that leaves it unmoved). The identity element of $$\mathbb{Z}/9\mathbb{Z}$$ is conventionally written $$0$$, but really the notation $$0$$ is short for the equivalence class of $$0$$ in $$\mathbb{Z}/9\mathbb{Z}$$, which is the coset $$0 + 9\mathbb{Z}$$. Sometimes people write $$\overline{0}$$ to emphasize that they mean the equivalence class.

You are correct that any homomorphism $$\phi: \mathbb{Z}/9\mathbb{Z} \to D_5$$ must send $$0 \mapsto \iota$$.

i'm not sure how it would work for Z/9 where i choose Z = 9.

What? "$$\mathbb{Z} = 9$$" just makes no sense.

In this case I get the identity element in Z/9 but in D₅ I get T⁴. Does this mean that Z/9 is not a homomorphism?

"Z/9 is not a homomorphism" is a true statement, but for the wrong reason: $$\mathbb{Z}/9\mathbb{Z}$$ is a group, not a homomorphism.

Here's a correct statement of what you seem to be saying: If you attempt to construct a homomorphism $$\phi: \mathbb{Z}/9\mathbb{Z} \to D_5$$ by sending $$1 \mapsto \rho$$, you find that on the one hand $$\phi(0) = \iota$$ because $$\phi$$ is a homomorphism, and on the other hand $$\phi(0) = \phi(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) = \rho^9 = \rho^4 \neq \iota$$. This contradiction implies that there is no homomorphism $$\phi: \mathbb{Z}/9\mathbb{Z} \to D_5$$ such that $$\phi(1) = \rho$$.

However, you haven't excluded all the maps $$\mathbb{Z}/9\mathbb{Z} \to D_5$$ that could possibly be homomorphisms yet.
(Hint: there is at least one map that is a homomorphism.)

If you use the facts about subgroups I mentioned above, you can solve the problem a little more efficiently than by enumerating all the possibilities for maps and proving one-by-one whether or not each is a homomorphism.

 

[nhartung]

I think I'm starting to understand this a little bit now. If i were to test for homomorphisms by enumerating all the possibilities for maps (which I realize would be a bad idea) would I be doing something like this:

Continuing with Z/9Z -> D₅:
Mapping 1 -> p² We find that phi(0) = phi(1+1+1+1+1+1+1+1+1) = p¹⁸ = p³ ≠ i.
Similarly we find that mapping 1 -> p³ and 1 -> p⁴ is not a homomorphism.
However when we map 1 -> i we find that phi(0) = phi(1+1+1+1+1+1+1+1+1) = i, meaning that phi(1) -> i is a homomorphism of Z/9 -> D5.

Is this correct? I hope so, now I need to figure out how to determine this by using Lagrange's theorm.

 

[ystael]

This is the right idea. However, there is something I glossed over a little bit in my previous post which you need to make sure you understand:

When one says "attempt to construct a homomorphism $$\phi: \mathbb{Z}/9\mathbb{Z} \to D_5$$ by sending $$1 \mapsto \iota$$ (this is the one that works), what that means is: to try to build a complete function from the domain to the range by taking the given data and extending it by using the homomorphism laws.

So the homomorphism $$\phi: \mathbb{Z}/9\mathbb{Z} \to D_5$$ given by $$\phi(1) = \iota$$ is the function that sends $$0 \mapsto \iota, 1 \mapsto \iota, 2 \mapsto \iota, \dots$$ -- that is, the constant function $$\phi(n) = \iota$$ for any $$n \in \mathbb{Z}/9\mathbb{Z}$$. (This is because a homomorphism $$\phi: G \to H$$ is determined completely by its effect on a set of generators of $$G$$, and $$\mathbb{Z}/9\mathbb{Z}$$ is a cyclic group with $$1$$ as a generator; $$\phi(2) = \phi(1)^2$$ and so on. You seem to understand this already, but it's worth explicitly pointing out why.)

You can't say that "$$\phi(1) = \iota$$ is a homomorphism of $$\mathbb{Z}/9\mathbb{Z}$$ to $$D_5$$" as you did above -- but it is a piece of data that can be successfully extended to form a homomorphism.