Show that the elements $x^iy^j$ such that $i=0,1,2,3$ and $j=0,1$ are distinct elements of $G$, and hence constitute all elements of $G$.

97 Views Asked by At

There exists a group $G$ of order $8$ having two generators $x,y:x^4=y^2=e$ and $xy=yx^3$.

I found that $G=\{e,y,x,xy,x^2,x^2y,x^3,x^3y\}$ But how to show that these elements are distinct? Is saying that $|G|=8$ enough? Or maybe I should check all 56 possibilities and show that at that case $|G|<8$? Also I don't understand how to show that we have all elements of $G$ by checking only for $0\leq i\leq 3$ and $0\leq j\leq 1$? Maybe there are other distinct element for some finite combination $xxyyxyyxyxyy...xyyx$?

I just want to know intuitively what I can do with groups and what I can't. So I need answers, and some of them may seem stupid to you, but I just need them and that's all.