Question:
If G is an abelian group with generators x and y of order 16 and 24 respectively, and $x^{2} = y^{3}$, what is the order of G?
My attempt at a solution:
Since the set of generators of the group is finite, and the order of each generator is given, we can combine the elements of the subgroups generated by x and y individually to get the total number of possible elements.
So, 16(24).
But then we must exclude repeats. I'm not sure where to go from here, working under the constraint that $x^{2} = y^{3}$. Any guidance as to how to take this constraint into consideration would be appreciated. Also, since the group is abelian would the total number of elements before this exclusion be 8(24)?
If $G$ is abelian and generated by $x$ and $y$, then any element can be written in the form $x^iy^j$ by commuting $x$ and $y$ as necessary. Since $x$ has order $16$ and $y$ has order $24$, we can also impose that $0\leq i \leq 15$ and $0 \leq j \leq 23$ which gives a total number of $16 \times 24$ elements.
If there were no other relations here, we would be done. However, we also know that $y^3=x^2$ which means that whenever we see a power of $y$ greater than $2$, we can turn some of those $y$'s into $x$'s and then reduce the power of $x$ to below $16$ again.
Hence any element can be written as $x^iy^j$ with $0\leq i \leq 15$ and $0\leq j \leq 2$ giving a total of $48$ elements and your group is (isomorphic to) $C_{16} \times C_3$, where $x=(1,0)$ and $y=(2,1)$.