How to solve the problem from Topics In Algebra Herstein?

Let $$G$$ be the dihedral group defined as the set of all formal symbols $$x^iy^j$$, $$i=0,1$$, $$j=0,1,\ldots,n-1$$, where $$x^2=e$$, $$y^n=e$$, $$xy=y^{-1}x$$. Prove

1. The subgroup $$N=\{e,y,y^2,\ldots,y^{n-1}\}$$ is normal in $$G$$.
2. That $$G/N\approx W$$, where $$W=\{1,-1\}$$ is the group under the multiplication of the real numbers.

I have solved (a) part .In part (b) we need to define homomorphic function.What i was thinking that after defining a homomorphic function if we prove that N is kernel then we are done .But i am unable to find a homomorphic function.

• I think your proposed homomorphism $\phi$ must sent every element of $N$ to $1$ (since $N$ is to be the kernel), so $\phi(y^k)=1$. Now? what must $\phi$ do to the $xy^k$? Commented Nov 6, 2018 at 19:21
• Your plan sounds reasonable. (Alternatively, you could try to show directly that the order of $G$ is $2n$, hence the order of $G/N$ is $2$.) For an explicit homomorphism function, consider $\phi(x^i y^j) = 1$ if $i$ is even, $-1$ if $i$ is odd. You will have to show that this is well-defined and a homomorphism.
– user169852
Commented Nov 6, 2018 at 19:21
• @Bungo I got your alternative hint.But i don't think we need to prove that G is a group for that (we can follow the assumption given in question)
– user584920
Commented Nov 7, 2018 at 4:55

Define$$\begin{array}{rccc}\varphi\colon&G&\longrightarrow&\{1,-1\}\\&x^iy^j&\mapsto&(-1)^i.\end{array}$$Prove that it is a group homomorphism. It is clear that $$\ker\varphi=N$$.
Hint - Define a similar map that Jose Carlos Santos has suggested in his answer from $$G/N$$ to $$W$$ . Since number of elements in $$G$$ is equal to $$2n$$ and number of elements in N is n this implies number of elements in $$G/N$$ is $$2$$ which is equal to number of elements in $$W$$.