I need to solved the following problem: Let $G$ be a group whose generators $n$ and $m$ satisfy the equations $n^2=e$, $m^3=e$ and $(nm)^2=(nm)(nm)=e$. Determine the multiplication table of $G$.
I'm try:
To determinate the multiplication table I try get all elements, so, in general, $G=\{n,e,m,m^2,nm,nm^2,nmn,nmnm,.....\}$ and with that information I try get the group using the equations $n^2=m^3=e$ but I don't know how I work with it. Someone could help me?
On
Note that as $n$ and $m$ are the generators of the group we have that each of the elements in $G$ is of the form $n^sm^t$, where $s = \{0,1\}$, $t = \{0,1,2\}$. This means that $G$ is of order $6$. Can you continue to make your multiplication table from this?
Also you can use that $nm = m^{-1}n^{-1}$ in order ot simplify some calculations, as well, as $n^{-1} = n$. Also use the fact that you should get all $6$ elements in each row and each column.
Since $m^3=e$, we get $m^{-1}=m^2$, and because $n^2=e$, we get $n^{-1}=n$. Then because $(nm)^2=e$, we get $nm=m^2n$. What is the significance of this rule?