How can I find subgroups of $\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_{12}$ isomorphic to Klein four-group? Good night... I'm trying to find all subgroups of $\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_{12}$ with excel, using $(a,b,c)$ and checking elements one by one and see if it is cyclic. Is there an easy way? How I can know how many subgroups are there? Is there a quick way to learn the generator of the group $\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_{12}$? How to determine if they are isomorphic to the Klein group? Thanks for your time!!!
 A: Klein four group is given by presentation $\langle a,b\, |\, a^2 = b^2 = e\rangle$, thus all you need to do is find two elements of order $2$ inside $\mathbb Z_2\times \mathbb Z_4\times \mathbb Z_{12}$. Note that for $(a,b,c)\in \mathbb Z_2\times \mathbb Z_4\times \mathbb Z_{12}$ $$2(a,b,c) = 0 \iff 2a = 2b = 2c = 0$$
so $a,b,c$ are of order $2$ or trivial. It follows that $(a,b,c)$ is of order $2$ if at least one of $a,b,c$ is of order $2$, and the rest are trivial. Thus, $a\in\{0,1\}$, $b\in\{0,2\}$, $c\in\{0,6\}$. This gives $2^3 - 1 = 7$ elements of order $2$ in $\mathbb Z_2\times \mathbb Z_4\times \mathbb Z_{12}$ and since we need to pick two of them to get Klein group, there should be $\binom 7 2=21$ subgroups isomorphic to Klein four group, but every subgroup is counted three times that way: if we pick two elements $x,y$ of order $2$, then $$\langle x,y\rangle = \langle x, x+y\rangle = \langle y, x+y\rangle = \{ 0, x, y, x+y\}$$ so in fact there are only $7$ subgroups isomorphic to Klein four group.
A: Klein four group is an abelian group but not cyclic
There is a subgroup Z2XZ2 is isomorphic to K4 but not to Z4 as it is cyclic
