I'm trying to compute the factor group of $(\Bbb Z_9\times\Bbb Z_{12})/\langle(4,6)\rangle$.

Question

I'm trying to compute the factor group of $(\Bbb Z_9\times\Bbb Z_{12})/\langle(4,6)\rangle$.

I know the order is $6$ but I'm not sure how to compute the cosets without writing out the $108$ terms of $\Bbb Z_9\times\Bbb Z_{12}$.

Is there a way to find the cosets without all the tedious work?

Answer 1

The factor group is an abelian group of order $6$ and so is cyclic of order $6$.

Indeed, the coset of $([1]_9,[1]_{12})$ has order $6$ in the factor group.

The elements of the factor group are the cosets of $([n]_9,[n]_{12})$ for $n=0,\dots,5$.