1. ## Question about presentation of cyclic group

I want to list the elements of $\displaystyle T=<a,b : a^6=1, b^2=a^3,bab^{-1}=a^{-1} >$

So from my understanding, $\displaystyle a^ia^j=a^{i+j} \ \ \ \forall 0 \leq i,j \leq 5$ with $\displaystyle a^{i+j}=a^{i+j-6} \ \ i+j \geq 6$

But I'm having trouble trying to make senses of the two other conditions, and how to properly translate them into the table.

Thank you!!!

2. Notice first that $\displaystyle b^4 = (b^2)^2 = (a^3)^2 = a^6 = 1$.

From the relation $\displaystyle bab^{-1} = a^{-1}$ it follows that $\displaystyle ba = a^{-1}b = a^5b$. Using that, in any product of a's and b's you can always push the a's to the left of the b's. For example $\displaystyle aba^2b = a(ba)ab = a(a^5b)ab = bab = a^5b^2$. In that way, you can express any element of the group as $\displaystyle a^ib^j$, with $\displaystyle 0\leqslant i\leqslant5$ and $\displaystyle 0\leqslant j\leqslant3$. That gives you 24 elements of the group, and you still have a fair amount of work to do if you want to write down the whole multiplication table.

3. and to complete Opalg's argument : also note that since $\displaystyle b^2=a^3$ and $\displaystyle b^3=a^3b,$ every element of $\displaystyle T$ can be written as $\displaystyle a^ib^j$ with $\displaystyle 0 \leq i \leq 5$ and $\displaystyle 0 \leq j \leq 1.$ it's easy to show that this

presentation is unique. so $\displaystyle T$ is a group of order $\displaystyle 12$ with elements $\displaystyle 1,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b.$ the group $\displaystyle T$ is called dicyclic group.