Normal Subgroups and Quotient Groups Let $G$ $(x^i y^j)$, $i = 0,1$, $j =0,1, 2, 3, \ldots, n-1$, where $(x^i y^j) = (x^{i'}y^{j'})$ iff $i = i'$, $j = j'$
$x^2 = y^n = e$, $n>2$
$xy = y^{-1} x$.
a) Find the form of the product $(x^i y^j)(x^k y^\ell)$ as 
      $(x^a y^b)$.
b) Using this, prove $G$ is a non-abelian group of order $2n$.
c) If $n$ is odd, prove the center of $G$ is $(e)$, while if $n$ is even, 
      the center of $G$ is larger than $(e)$.
 A: (a) First note that if $0\le k<n$, then $(y^k)(y^{n-k})=y^n=e$, so $(y^k)^{-1}=y^{n-k}$. In particular, $y^{-1}=y^{n-1}$. We also need to generalize the third assumption, that $xy=y^{-1}x$: prove by induction on $k$ that $xy^k=(y^{-1})^kx=y^{n-k}x$ for $k=0,\dots,n-1$.
Now we’ll look at $(x^iy^j)(x^ky^\ell)$. 


*

*If $j=0$, this is simply $x^{(i+k)\bmod 2}y^\ell$; here we use the fact that $x^2=e$.  

*If $k=0$, it’s $x^iy^{(j+\ell)\bmod n}$; here we use the fact that $y^n=e$.  

*If $j\ne 0$ and $k=1$, we need to turn $y^jx^k=y^jx$ into a product with the power of $x$ first. Let $r=n-j$; then $y^jx=y^{n-r}x=xy^r=xy^{n-j}$ by the generalization of the third assumption that I mentioned above. Then $$(x^iy^j)(xy^\ell)=x^ixy^{n-j}y^\ell=x^{(i+1)\bmod 2}y^{(n-j+\ell)\bmod n}=x^{(i+1)\bmod 2}y^{(\ell-j)\bmod n}\;.$$


If you look carefully at these, you’ll see that they can be combined: the formula
$$(x^iy^j)(x^ky^\ell)=\begin{cases}
x^iy^{(j+\ell)\bmod n},&\text{if }k=0\\
x^{(i+k)\bmod 2}y^{(\ell-j)\bmod n},&\text{if }k=1
\end{cases}\tag{1}$$
covers all three cases correctly.
(b) Clearly there are $2n$ formal symbols $x^iy^j$ with $i\in\{0,1\}$ and $j\in\{0,\dots,n-1\}$, and they’ve been defined to be distinct, so it only remains to show that they form a group under the operation defined in $(1)$: you must show that the operation is associative, that it has an identity, and that it has inverses. I’ll leave the details to you, but the identity is $e=x^0y^0$, and you can easily use $(1)$ to figure out that the inverse of $x^0y^j$ is $x^0y^{n-j}$, and the inverse of $x^1y^j$ is $x^1y^j$. Finally, to see that $G$ is non-Abelian, observe that $(x^1y^0)(x^0y^1)=x^1y^1$, while $(x^0y^1)(x^1y^0)=x^1y^{n-1}\ne x^1y^1$ (provided that $n>2$).
(c) If $n=2m$, show that $x^1y^m$ is in the centre of $G$, but if $n$ is odd, the only elements of $G$ that commute with $x^1y^0$ are itself and $e$.
A: Hint: show that $\langle y\rangle$ is normal in $G$, and $\langle x\rangle\cap\langle y\rangle=1$.
A: Hint: See if you can link the laws together to produce some new information. Note that $e$ is a symbol for the identity. $x^2=y^n=e$ is clearly a useful situation to corner things into, but what do you want to corner? Try a special case of $(x^ix^j)(x^kx^l)$.
