1
$\begingroup$

I am trying to find all subgroups of order 4 in Z4 x Z4. I have:

  • $\{0\} \times Z_4 = \langle(0,1)\rangle$
  • $Z_4 \times \{0\} = \langle(1,0)\rangle$
  • $\langle(1,1)\rangle$
  • $\langle(0,2)\rangle$

Have I missed any?

$\endgroup$
  • $\begingroup$ What about $\;\langle (1,0)\rangle\;,\;\;\langle (0,1)\rangle$ ? BTW, $\;2\cdot (0,2)=(0,0)\implies \langle (0,2)\rangle\;$ has order two. $\endgroup$ – DonAntonio Feb 23 '14 at 20:02
  • $\begingroup$ isn't that first one of order 8? $\endgroup$ – Paul Malinowski Feb 23 '14 at 20:04
  • $\begingroup$ It can't be @Paul as the group's exponent is $\;4\;$ ... $\endgroup$ – DonAntonio Feb 23 '14 at 20:07
  • $\begingroup$ @DonAntonio what do you mean by $\;\langle (1,0)\rangle\;,\;\;\langle (0,1)\rangle$? If you mean $\langle (1,0)\rangle$ and $\langle (0,1)\rangle$, both of those are listed. $\endgroup$ – Omnomnomnom Feb 23 '14 at 20:08
  • $\begingroup$ Either I missed those (as usual), or were edited before 5 minutes passed, @Omnomnomnom...thanks. $\endgroup$ – DonAntonio Feb 23 '14 at 20:09
1
$\begingroup$

Note that $\langle (0,2) \rangle$ is not of order $4$. However, $\langle (0,2),(2,0) \rangle$ is.

Note additionally that you're missing $\langle(1,3)\rangle,\langle(1,2)\rangle$, and $\langle(2,1)\rangle$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.