Find injective homomorphism from D_2n to S_n

[Infrared]

Every element of D_{2n} is either of the form r^i or sr^i for some i. The first case maps to \tau^i, which is addition by i modulo n, so can only be the identity if i is a multiple of n, meaning that r^i=1. The second case maps to \sigma\tau^i. This permutation takes 1 to -i, so could only possibly be trivial if i=n-1. But in that case it doesn't fix n.

I think I gave you too much help. In the future, maybe try more on your own.

 

[Mr Davis 97]

Every element of D_{2n} is either of the form r^i or sr^i for some i. The first case maps to \tau^i, which is addition by i modulo n, so can only be the identity if i is a multiple of n, meaning that r^i=1. The second case maps to \sigma\tau^i. This permutation takes 1 to -i, so could only possibly be trivial if i=n-1. But in that case it doesn't fix n.

I think I gave you too much help. In the future, maybe try more on your own.

I actually have one more question. I'm trying to find the cycle type of each element in the image of the homomorphism. I think I worked out correctly that the cycle type of $\tau^i$ is $\displaystyle \frac{n}{\operatorname{gcd} (i,n)}$, repeated as many times as need to get $n$. For example, if $n = 16$ then $\tau^4$ has cycle type 4,4,4,4. Is this right?

Also, I need a bit of help figuring out what the cycle type for $f(sr^i)$ would be for any $i$.

 

[Infrared]

Why do you need to find the cycle type?
In any event, yes, that expression is correct. The sr^i are all reflections, so will map to a product of n/2 disjoint transpositions if the line of reflection is through an edge, and n/2-1 disjoint transpositions if the line of reflection is through opposite vertices.

 

[Mr Davis 97]

Why do you need to find the cycle type?
In any event, yes, that expression is correct. The sr^i are all reflections, so will map to a product of n/2 disjoint transpositions if the line of reflection is through an edge, and n/2-1 disjoint transpositions if the line of reflection is through opposite vertices.

It's just an additional aspect of the problem. So I think this is all the help I need. Just one more thing. Do I have to do all of this again for the case when $n$ is odd? That would seem to be very tedious...

 

[Infrared]

Most of the work is very similar/identical, so you can probably get away with saying something like "by arguing as before, we have...".

 

[Infrared]

I think we went overkill on detail. For the purpose of the problem, you could probably just say: labeling the vertices associates to each element of D_{2n} a permutation in S_n. This association is a homomorphism because multiplication in D_{2n} corresponds to composing permutations. This association is injective because only the identity element of D_{2n} fixes all vertices. Then write down the map explicitly as we did.

 

[Mr Davis 97]

I think we went overkill on detail. For the purpose of the problem, you could probably just say: labeling the vertices associates to each element of D_{2n} a permutation in S_n. This association is a homomorphism because multiplication in D_{2n} corresponds to composing permutations. This association is injective because only the identity element of D_{2n} fixes all vertices. Then write down the map explicitly as we did.

You're probably right, and I'll take that into account when writing my solution. There's one more thing I'm curious about. My problem says "Find, with justification, an injective group homomorphism from $D_{2n}$ to $S_n$." This makes it seem like I have to find one that works for all cases of $n$. But I had to split it into two cases to get two different homomorphisms... Is the problem just phrased poorly?