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Find all subgroups of order $4$ in $Z_4 \oplus Z_4.$

My attempt:

We begin to find all elements of order $4$ in $Z_4 \oplus Z_4.$ First attempt is to find all the cyclic subgroups of order $4.$ We want lcm$\big(|g_1|,|g_2|\big)=4$ where $g_1,g_2\in Z_4.$ So, the distinct generators of cyclic groups of order $4$ would be $\langle(0,1)\rangle,\langle(1,0)\rangle,\langle(2,1)\rangle ,\langle(1,2)\rangle ,\langle(1,1)\rangle \&\langle(1,3)\rangle.$

A non-cyclic group of order $4$ should be of the form $\{e,a,b,ab\}$ and all the non-identity elements should have order $2.$ So, we want lcm$\big(|g_1|,|g_2|\big)=2$ where $g_1,g_2\in Z_4.$ Therefore the elements could be $(0,2),(2,0)\& (2,2).$ So the group $\{(0,0),(0,2),(2,0),(2,2)\}$ is also a subgroup of order $4$.

Therefore there are $7$ groups of order $4.$

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  • $\begingroup$ Why $<(1,3)>$ but not $<(3,1)>$ ? Also, I think you forgot $<(3,3)>$. $\endgroup$ – paf Jun 9 '16 at 16:44
  • $\begingroup$ @paf they will generate the same group. But I agree it could be explained more clearly. $\endgroup$ – quid Jun 9 '16 at 16:52
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    $\begingroup$ Your math is correct! (By the way, it looks better to use \langle and \rangle as delimiters rather than the less-than and greater-than signs: compare $\langle (0,1) \rangle$ to $<(0,1)>$.) $\endgroup$ – Greg Martin Jun 9 '16 at 17:09
  • $\begingroup$ noted @GregMartin $\endgroup$ – Bijesh K.S Jun 9 '16 at 17:34
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Yes, you are correct!

Here is a relevant OEIS sequence entry. It agrees with your calculation.

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