Find all subgroups of $\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}$ isomorphic to the Klein $4$-group.

This answer addresses the question in the title, not the many all along the post.

The elements of order $2$ of $G:=\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_4$ are:

\begin{alignat}{1} &a_1=(0,0,2), \space a_2=(0,1,2), \space a_3=(1,0,2), \space a_4=(1,1,2) \\ &a_5=(0,1,0), \space a_6=(1,0,0), \space a_7=(1,1,0) \\ \tag 1 \end{alignat}

Any candidate subgroup of $G$, say $K$, isomorphic to the Klein group must be made of:

1. the unit $1_G:=(0,0,0)$;
2. any pair $a_i, a_j$;
3. the element $a_i+a_j$; in fact: $a_i+a_j\in K$ by closure, and $a_i+a_j\ne 1,a_i,a_j$.

Is, for every $1\le i<j\le 7$, the subset $K_{ij}:=\{1_G,a_i,a_j,a_i+a_j\}$ indeed a subgroup of $G$? It's enough to prove the closure:

• $a_i^2=a_j^2=(a_i+a_j)^2=1_G$
• $a_i+(a_i+a_j)=(a_i+a_i)+a_j=a_j$
• $a_j+(a_i+a_j)=a_j+(a_j+a_i)=(a_j+a_j)+a_i=a_i$

So, indeed $K_{ij}\le G$ and $K_{ij}\cong\mathbb{Z}_2\times\mathbb{Z}_2$.

Now, if we denote $a_k:=a_i+a_j$, then $a_k+a_i=a_j$ and $a_k+a_j=a_i$. So:

1. if $k<i<j$, then $K_{ij}=K_{ki}=K_{kj}$;
2. if $i<k<j$, then $K_{ij}=K_{ik}=K_{kj}$;
3. if $i<j<k$, then $K_{ij}=K_{ik}=K_{jk}$.

Therefore, the number of (distinct) subgroups of $G$ isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ is:

\begin{alignat}{1} n_K &= \frac{1}{3}\cdot|\{K_{ij}, \space1\le i<j\le 7\}| \\ &= \frac{1}{3}\cdot\frac{7\cdot 7-7}{2} \\ &= 7 \\ \tag 2 \end{alignat}

Explicitly, according to the labelling $(1)$:

\begin{alignat}{1} K_{12} &= \{1_G,a_1,a_2,a_5\}\space (=K_{15}=K_{25}) \\ K_{13} &= \{1_G,a_1,a_3,a_6\}\space (=K_{16}=K_{36}) \\ K_{14} &= \{1_G,a_1,a_4,a_7\}\space (=K_{17}=K_{47}) \\ K_{23} &= \{1_G,a_2,a_3,a_7\}\space (=K_{27}=K_{37}) \\ K_{24} &= \{1_G,a_2,a_4,a_6\}\space (=K_{26}=K_{46}) \\ K_{34} &= \{1_G,a_3,a_4,a_5\}\space (=K_{35}=K_{45}) \\ K_{56} &= \{1_G,a_5,a_6,a_7\}\space (=K_{57}=K_{67}) \\ \tag 3 \end{alignat}