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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.

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    $\begingroup$ 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 $xy^k$? $\endgroup$ Commented Nov 6, 2018 at 19:21
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    $\begingroup$ 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. $\endgroup$
    – user169852
    Commented Nov 6, 2018 at 19:21
  • $\begingroup$ @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) $\endgroup$
    – user584920
    Commented Nov 7, 2018 at 4:55

2 Answers 2

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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 $\ker\varphi=N$.

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    $\begingroup$ Just from little curiosity - I will try to solve the problem by using the homomorphic map you suggested.It is possible to have different homomorphic map's for solving the problem(As math.stackexchange.com/users/436618/lord-shark-the-unknown has commented to question) ? $\endgroup$
    – user584920
    Commented Nov 6, 2018 at 19:37
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    $\begingroup$ No. The map that I mention is the only one that will do. And Lord Shark the Unknown never suggested otherwise. $\endgroup$ Commented Nov 6, 2018 at 19:40
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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 $W$.

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