0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.