# Let $G$ be a group defined as all formal symbols $x^{i}y^{j}$

Let $G$ be a group defined as all formal symbols $x^{i}y^{j}$, $i=0,i,j=0,1,2,...,n-1$ where we assume

$x^{i}y^{j}=x^{i'}y^{j'}$ if and only if $i=i',j=j'$

$x^2=y^n=e$ for $n \gt 2$

$xy=y^{-1}x$.

(a)Find the form of the product $(x^{i}y^j)(x^ky^l)$ as $x^{\alpha}y^{\beta}$.

(b)Using this prove that $G$ is a non-abelian group of order $2n$.

(c)If $n$ is odd, prove that Centre of $G$ is $(e)$, while if $n$ is even the centre of $G$ is larger than $(e)$.

For (a) we will continue till $i-2n$ and $k-2n$ becomes equal to $2$ or less than that. Either way we will end up at $y^jxy^l$, or $xy^jxy^l$ or $xy^{j+l}$ or $y^{j+l}$.

Then $y^jxy^l=y^{j-1}yxy^l=y^{j-1}xy^{-1}y^l=y^{j-2}yxy^{-1}y^l=y^{j-2}x(y^{-1})^2y^l=x(y^{-1})^jy^l=xy^{l-j}$.

Then $xy^jxy^l$ beomes $y^{l-j}$. The rest cases are straightforward.

For (b) let's look at $\{e,x,y,y^2,...y^{n-1},xy,xy^2,xy^3,..,xy^{n-1}\}$. Now $xy^{n-k}.y^{k}x=e=y^{k}x.xy^{n-k}$.Besides $y^kx=xy^{n-k}$. All I need to show that the elements are distinct. Now If $xy^i=xy^j$, by the first assumption $i=j$. For the non-abelian thing if $xy=yx$ then by the third assumption $y^{-1}x=yx\implies y^2=e$ not possible by the second assumption.

For (c) if $n=2k$ then $y^k=y^{-k}$. Then $(xy^i)y^k=xy^{i+k}=xy^{i-k}=(y^k)(xy^{i})$. So $y^k$ lies in the centre of $G$.

If $n=2k-1$ , and there exists an element $p$ such that $gp=pg$ for all $g$ In $G$ then from here I am stuck.

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Your first lines are inconsistent. If $x^iy^j=x^{i'}y^{j'}$ if and only if $i=i'$ and $j=j'$ then certainly you do not have $x^2=y^n$. –  Jérémy Blanc Apr 7 at 9:01
dude these are the assumptions. I didn't create them. @ jeremy –  tattwamasi amrutam Apr 7 at 11:47
So the answer is "this group does not exist". –  Jérémy Blanc Apr 7 at 12:35
this group is called Dihedral group and it exists. –  tattwamasi amrutam Apr 7 at 13:13
What you probably want to say is that your group is generated by two symbols $x,y$ and that you take the relations $x^2=y^n=e$, $xy=y^{-1}x$ and yes in this case you get the Dihedral group, but not with the assertion "$x^iy^j=x^{i'}y^{j'}$ if and only if $i=i'$ and $j=j'$". Choose for example $i=2$, $j=0$, $i'=0$, $j=n$. –  Jérémy Blanc Apr 7 at 13:46