Say we have the Group $(\Bbb Z_3 \times \Bbb Z_4,+)$, this group has the following elements;

$\{(0,0)(0,1)(0,2)(0,3)(1,0)(1,1)(1,2)(1,3)(2,0)(2,1)(2,2)(2,3)\}$ and so is a group of order 12. Therefore ( I believe) that the subgroups are found by looking for groups of order 1,2,3,4,6,12.

We can also note that $|\langle g \rangle|=o(g)$(i.e the order of the element g)

$\therefore \langle(0,1)\rangle:=\{(0,1)(0,2)(0,3),(0,0)\}$ is a subgroup as $o((0,1))=4$ and $|\langle(0,1)\rangle|=4$

$\langle(2,3)\rangle:=\{(2,3),(1,2),(0,1),(2,0),(1,3),(0,2),(2,1),(1,0),(0,3),(2,2)(1,1)(0,0)\}$ has order 12 and so is also a subgroup.

When I was going though the list of elements in $\Bbb Z_3 \times \Bbb Z_4$ It seemed like they all generated a group of order 1,2,3,4,6 or 12, Is this true ?

Also if my method for computing subgroups is correct, is there any other ways to find subgroups apart from using this generator technique ?

  • $\begingroup$ Your group is isomorphic to $\mathbb{Z}_{12}$, see the links on MSE, or this homework solution. $\endgroup$ – Dietrich Burde May 13 '18 at 19:19
  • $\begingroup$ Use $\langle\rangle$ for $\langle\rangle$. $\endgroup$ – Shaun May 13 '18 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.