Hello,

Let $G = \langle a,b,c \; | \; a^2 = b^3 = c^4=1, cbc^{-1}=a \rangle$ find the order of $G$. There are many such possible values find them all.

The condition that $c^4 =1$ means that the order of $c$ divides 4 so the order is either 2 or 4 (we ignore 1 because we don't want $c$ to be the identity element.)

So I began by assuming the order of $c$ was 2. I don't recognize the group from it's presentation so I decided to try working it out as a free group, you take all possible combinations of distinct elements, the relations should collapse some of the possibilities down, and then you can just count distinct elements.

The first problem I ran into is that I don't really know what to do with $cbc^{-1} = a$ other than writing $cb=ac$ and $(cb)^{-1} = (ac)^{-1}$. It looks like it says that the subgroup generated by $b$ is conjugate to the subgroup generated by $a$ but I don't really know what I can do with that information (if it is correct).

The second problem is that the list of distinct elements is getting pretty big quickly. I'm up to 45+ elements and I've only multiplied $b^2$ by distinct elements. When I multiply by $c$ the list should double in size assuming nothing collapses.

Is there a better/faster way to do this? This process should terminate but it will take way too long to be practical. Calculating the case where the order of the subgroup generated by $c$ is 4 would take a very long time by hand.