Compute the factor group $(\mathbb{Z_6}\times \mathbb{Z_4})/\langle(3,2)\rangle$ and $\langle(3,2)\rangle=\{(0,0), (3,2)\}$. Here is what I have so far. I know I have 12 left cosets which are. $$(1,0)+ \langle(3,2)\rangle =\{(1,0),(4,2)\}$$ $$(2,0)+ \langle(3,2)\rangle=\{(2,0),(5,2)\}$$ $$(3,0)+ \langle(3,2)\rangle=\{(3,0),(0,2)\}$$ $$(4,0)+ \langle(3,2)\rangle =\{(4,0),(1,2)\}$$ $$(5,0)+ \langle(3,2)\rangle =\{(5,0),(2,2)\}$$ $$(0,1)+ \langle(3,2)\rangle =\{(0,1),(3,3)\}$$ $$(0,3)+ \langle(3,2)\rangle =\{(0,3),(3,1)\}$$ $$(1,1)+ \langle(3,2)\rangle =\{(1,1),(4,3)\}$$ $$(1,3)+ \langle(3,2)\rangle =\{(1,3),(4,1)\}$$ $$(2,1)+ \langle(3,2)\rangle =\{(2,1),(5,3)\}$$ $$(2,3)+ \langle(3,2)\rangle =\{(2,3),(5,1)\}$$ Now I have to find a group that this is isomorphic to which I am having trouble with. I think its isomorphic to $\mathbb{Z_{12}}$ but I can't seem to find a generator for the cossets. Thanks

  • $\begingroup$ I've changed things like $<(3,2)>$ to $\langle(3,2)\rangle$ in the first several instances. I'll leave it to others to clean up the rest of this. TeX and LaTeX were created for a purpose. The purpose was NOT to enable people to feel that they're restricted to the characters on a typewriter keyboard. $\endgroup$ Oct 18, 2013 at 3:40
  • $\begingroup$ oh ok i changed it. $\endgroup$
    – user60887
    Oct 18, 2013 at 3:42
  • $\begingroup$ I got it I think it was $(2,1)+ \langle(3,2)\rangle$ that has order 12. $\endgroup$
    – user60887
    Oct 18, 2013 at 3:52

1 Answer 1


The element $(2,1)+\langle(3,2)\rangle$ is indeed of order 12, precisely because $(2,1)$ is of order 12 in $\mathbb{Z}_6 \times \mathbb{Z}_4$, and the subgroup in $\mathbb{Z}_6 \times \mathbb{Z}_4$ generated by $(2,1)$ does not contain $(3,2)$. This can be seen by observing that the 3 in the left entry of $(3,2)$ cannot be generated by the 2 in the left entry of $(2,1)$.


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