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Find all sub groups of order 4 in $\mathbf{Z}_4 \oplus \mathbf{Z}_4$.

Solution : $\mathbf{Z}_4 =\{0,1,2,3\}$

$O(1) = O(3) = 4$, $O(0) = 1$, $O(2) = 2$

Hence, I found the subgroups of order 4 as follows:

$\langle 1,0 \rangle,\langle 0,1 \rangle,\langle 0,\mathbf{Z}_4 \rangle,\langle \mathbf{Z}_4, 0 \rangle,\langle 1,1 \rangle,\langle 1,3 \rangle,\langle 3,1 \rangle,\langle 3,1 \rangle,\langle 1,2 \rangle,\langle 2,1 \rangle$.

So, all of these are cyclic sub groups. How do we know that there are only cyclic subgroups in here and not some non-cyclic groups as well?

$\mathbf{Z}_4 \oplus \mathbf{Z}_4$ is not cyclic since $\gcd(|\mathbf{Z}_4|,|\mathbf{Z}_4|) \neq 1$, hence there can be non-cyclic subgroups of order 4 as well. How do we make sure apart from directly computing?

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Hint :

  • How does a non cyclic group of order $4$ look like?
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  • $\begingroup$ Suppose a group $G$ of order $4$ is non cyclic. Then, $\nexists$ any $g \in G ~ s.t.~ g^4 = e. =>$ order of every element in $G$ must be $2$. Hence, $G = \{e,a,b,ab\}~s.t.~ O(a) = O(b) = O(ab) = 2$ . Then, $G \thickapprox Z_2 \bigoplus Z_2$ $\endgroup$ – MathMan Apr 17 '14 at 9:57
  • $\begingroup$ So, what do you think you have to do now??? check for two elements of order $2$ whose product is also of order $2$...Can you do that? $\endgroup$ – user87543 Apr 17 '14 at 9:58
  • $\begingroup$ I am asking for elements in $\mathbb{Z}_4\times \mathbb{Z}_4$ where as you are saying about elements in $\mathbb{Z}_4$.. Can you be more careful? $\endgroup$ – user87543 Apr 17 '14 at 10:04
  • $\begingroup$ So, we have $(2,0)$ and $(0,2)$ as the two elements of order $2$. Their product is $(2,0)(0,2) = (2+0, 0+2) = (2,2)$ which is of order $lcm (2,2) = 2$ $\endgroup$ – MathMan Apr 17 '14 at 10:07
  • $\begingroup$ $(2,0)(0,2) = (2+0, 0+2) = (2,2)$ which is of order $lcm (2,2) = 2$ ? we have two elements whose product has order $2$ ? $\endgroup$ – MathMan Apr 17 '14 at 10:10

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