Homomorphism of dihedral group

Hey I've been working on this question,

How that the following is a homomorphism

From what it looks like it isn't a homomorphism but I'm not sure how to evaluate the last line,

It would be easy if the group commuted..

Does anyone know how to evaluate it?

I also have another question which I solved numerically but I'm not sure how to show it algebraically,

Would anyone know how to show

Or would it be sufficient to just know that its not true?

Re: Homomorphism of dihedral group

The dihedral group consists of rotations and reflections; the rotations form a cyclic subgroup of order An element of can be written as where is any reflection and or

A rotation generally does not commute with a reflection; however, any product of a reflection followed by a rotation is equal to a product of a rotation followed by another reflection. In other words, for some other Hope this helps.

For your second question, if you take and then

Re: Homomorphism of dihedral group

Thanks Sylvia it definitely helped

Re: Homomorphism of dihedral group

Sorry to bug you again,

I thought I had the question down but then I realised I made an assumption that the group binary operation was multiplication,

The question was show the following is a homomorphism

When I did this step I assumed multiplication,

but I now I'm paranoid that it should be addition,

Since in class we mainly use the integers under an addition modulo rather then multiplication, we only do multiplication on integers co-prime to the dimension

If n = 4 in this case then would the operation be addition modulo ten then 0 = 2+2=4 with a=b=6 like you said before?

I read online somewhere that "Reduction modulo n is a homomorphism" is that true?

I found some cases by mucking around in mathematica where its not satisfied under addition, if n=5 a=11 b=8,

I should of posted the question up in the first place haha

Sorry again for pestering you once more :p

Re: Homomorphism of dihedral group

Re: Homomorphism of dihedral group

Ahh so it is addition!

phi(a+b) = Rem[(a+b)/n] = Rem[a/n]+Rem[b/n] = phi(a) + phi(b) !

Thanks again Sylvia!

Re: Homomorphism of dihedral group

here is a (somewhat more) formal proof. suppose a = a' (mod n), and b = b' (mod n). then a+b = a'+b' (mod n).

since a = a' (mod n), by definition a - a' = kn, for some integer k. we can re-write this as a = a' + kn.

similarly, since b = b' (mod n), then b - b' = mn, for some integer m. we can re-write this as b = b' + mn.

therefore: a + b = (a' + kn) + (b' + mn) = a' + b' + (k + m)n. thus (a + b) - (a' + b') = (k + m)n, so

a + b = a' + b' (mod n) (since k+m is an integer whenever k and m are).

in particular, if [a] is the unique element of Z congruent to a mod n between 0 and n-1, then [a] + [b] = [a + b],

which makes the map a-->[a] a homomorphism from Z to Zn: φ(a + b) = [a + b] = [a] + [b] = φ(a) + φ(b).

Re: Homomorphism of dihedral group