Find all subgroups of $\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}$ isomorphic to the Klein $4$-group. John Fraleigh. A First Course in Abstract Algebra (7 edn 2002). p 110. Exercise 11.12.
I  have 8 questions on this Yahoo Answer. I know

*

*$|\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}| = 2\cdot2\cdot4$.


*$order(\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}) = \mathrm{lcm}(2, 2, 4) = 4.$ Hence by Lagrange's Theorem, $|\text{subgroup}| = \text{$1$, $2$ or $4$}$.

You need to simply find two (distinct) elements of order 2.


*

*$\color{red}{\text{ Why simply two distinct elements of order 2 ?  }  } $

Their sum will also be of order 2.  What are the elements of order 2? $a = (1,0,0), \;
b = (0,1,0), \;
c = (0,0,2) $
These are the "basic" ones.



*$ \color{red}{\text{ How do you deduce these elements? Surely not by trial and error? }  } $


*$ \color{red}{\text{ How do you envisage there are more elements of order 2?}  } $
As per p. 61 Theorem 6.6 in the aforementioned Fraleigh textbook, any subgroup of a cyclic group is cyclic.
Hence by reason of my question 4 above] the sum of these 2 distinct elements will also have order 2.

Those are the "basic" ones. Now take all possible sums. $a+b = (1,1,0), \; a+c = (1,0,2), \; b+c = (0,1,2), \; a+b+c = (1,1,2)$.
These are the only elements of order 2. We will sort these out by the number of $\{a,b,c\}$ that are in the subgroup of $\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}$ of order 2.



*$ \color{red}{\text{ Why do we need to sort? }  } $

Notice that you can't have all three, since all three would give you the whole 7 elements of order 2 $ > |V_4| = 4$.

Possibility 1 of 3. Any two of $\{a,b,c\}$ are in the subgroup.   Then these 2 add to determine the third.

If any pair of a,b,c are in it, it determines the third. So in this case, you are done -- there are $\binom{3}{2}$ ways to do that:
\begin{aligned}
\langle a , b \rangle &= \{ e, a, b, a+b \}\\
&= \{ (0,0,0) , (1,0,0) , (0,1,0) , (1,1,0) \} \\
\langle a , c \rangle &= \{ e, a, c, a+c \}\\
&= \{ (0,0,0) , (1,0,0) , (0,0,2) , (1,0,2) \} \\
\langle b , c \rangle &= \{ e, b, c, b+c \}\\
&= \{ (0,0,0) , (0,1,0) , (0,0,2) , (0,1,2) \}
\end{aligned}

Possibility 2 of 3. Only one of $\{a,b,c\}$ is in the subgroup.

Finally, what about subgroups that contain exactly one of $\{a,b,c\}$ ?  If it contains a, for example, then it cannot contain a+b, or a+c, since that would give a+(a+b)=b or a+(a+c)=c.

I rewrote this Possibility 2 more clearly. Suppose we're working with $a$. Then why can't this subgroup  contain $a+b$ or $a+c$? If it did, then by group closure, $a + (a + b) = b$ or $a + (a + c) = c$ are in the subgroup.
This contradicts our supposition that only one of $\{a,b,c\}$ is in the subgroup.

So it must contain $\{b + c, a + b + c\}$.



*$ \color{red}{\text{ Why must it "contain $\{b + c, a + b + c\}$"?}}$

This gives our other three groups:
$$\begin{aligned}
\langle a , a+b+c \rangle &= \{ e , a , b+c , a+b+c \}\\
&= \{ (0,0,0) , (1,0,0) , (0,1,2) , (1,1,2) \} \\
\langle a , a+b+c \rangle &= \{ e , b , a+c , a+b+c \}\\ 
&= \{ (0,0,0) , (0,1,0) , (1,0,2) , (1,1,2) \} \\
\langle a , a+b+c \rangle &= \{ e , c , a+b , a+b+c \}\\ 
&= \{ (0,0,0) , (0,0,2) , (1,1,0) , (1,1,2) \}
\end{aligned}$$

Possibility 3 of 3 . None of $\{a,b,c\}$ is in the subgroup.

In that case, if you have $a+b+c$, you are out of luck -- any third element will be something like $a+b$ and $(a+b+c)+(a+b)=c.$



*$ \color{red}{ \text{ I can't grasp what this Possibility 3 is saying. Why "out of luck"? How does $a + b + c \implies$ "any third element will be something like  $a+b$ and $(a+b+c)+(a+b)=c$}"?  } $

Done.





If you want a more "advanced" explanation of this, what's neat is that you have this:

*

*There are 7 elements of order 2.

*A subgroup contains exactly 3 of them.

*Any two of these elements determines the third (and thus the entire subgroup).



Now notice the axioms of a finite projective plane of order 7:

*

*There are 7 points.

*Any line contains exactly 3 points.

*Any two points determine a unique line.



Here, just make the correspondence
point $\longleftrightarrow$ element of order 2
line $\longleftrightarrow V_4$ subgroup


But a finite projective geometry of order 7 has 7 lines. That's the number of lines (subgroups) we already found by the "boring" method. But this is much cooler! Here is a diagram of this geometry:



It is called the "Fano plane." I have labeled the points with their corresponding elements in your group. Notice how the three lines of the first type are the sides of the triangle, the three of the last type go from vertex to midpoint, and the "exceptional" line is actually a "circle" in shape (here "line" doesn't mean "straight line") through the three midpoints.





I got an email from the asker, asking for clarification...
First, we can ignore the elements of order 4.



*$ \color{red}{\text{ Why can we "ignore the elements of order 4"?}}$

We are looking at the subgroup of Z2 x Z2 x Z4 which consists of elements of order 2. Because the group is [A]belian, this is a legitimate subgroup. Call it H.


Then the set ${a,b,c}$ is a generating set of H. Further, H has order 8. It has 7 nonzero elements, and they will all be order 2 by definition.
All Klein groups are subgroups of H, obviously.



*$\color{red}{\text{ This isn't obvious to me. How are all Klein groups subgroups of $H = {a, b, c}$?}}$

Then you just use the above logic to figure out how many subgroups of $H$ there are. The above logic tells us how. You ask why you can't have $a,b,c$ all in the same Klein group, but ${a,b,c}$ generates $H$, so if they are in the same subgroup, that subgroup is too big to be a Klein group.


The fact that $H$ forms a projective geometry is awesome, but if you are having trouble grasping the basic definitions, you can ignore that for now...

 A: 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:

*

*the unit $1_G:=(0,0,0)$;

*any pair $a_i, a_j$;

*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:

*

*if $k<i<j$, then $K_{ij}=K_{ki}=K_{kj}$;

*if $i<k<j$, then $K_{ij}=K_{ik}=K_{kj}$;

*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}
A: 

*

*Why simply two distinct elements of order 2?


The Klein $4$-group consists of three elements of order $2$ and the identity, and it is up to isomorphism the only group with these properties. As the sentence quoted below your question explains, once you've chosen two such elements, you get the third as sum of the two you've chosen. It is not hard to show that for any two order-$2$ elements $x$ and $y$, the set $\{0,x,y,x+y\}$ indeed is a subgroup isomorphic to the Klein $4$-group, and of course any such group has to be of that form.



*How do you deduce these elements? Surely not by trial and error?


Well, for any product $G_1\times G_2\times\ldots\times G_n$, we have the natural subgroups $G_1\times\{e_2\}\times\ldots\times\{e_n\}$, $\{e_1\}\times G_2\times\ldots\times\{e_n\}$, …, $\{e_1\}\times\{e_2\}\times\ldots\times G_n$, where $e_i$ is the neutral element of $G_i$.
Note that those subgroups are disjoint in the sense that for any two of them, the only element shared is the neutral element. Moreover, each elements of the product group can be written uniquely as product whose factors are one element from each of those subgroups.
In your case, the neutral elements of the factors are $0$, and each of the groups has exactly one element of order $2$; it's $1$ for $\mathbb Z_2$ and $2$ for $\mathbb Z_4$. Thus by taking the elements or order two from each of the factors and “padding” them with zeroes we get the three elements given: $(1,0,0)$, $(0,1,0)$ and $(0,0,2)$.



*How do you envisage there are more elements of order 2?


As mentioned above, the ones ion this list are from very specific subgroups. Also as mentioned before, the sum of two elements of order $2$ gives another element of order $2$. And if those two elements are from disjoint subgroups, their sum clearly is in neither of the subgroups, so they are not in that list.



*Why do we need to sort?


That's a misunderstanding. The sentence doesn't say “we will sort these” but “we will sort these out”. To sort out doesn't mean to sort them, but rather to select them on some criterion. In this case, based on the criterion on how many of those listed elements are in the subgroup.
Note also that this is not a need, but a way to simplify your work while making sure you don't miss any case. In principle you could manually check every single of the $35$ different sets of three order-two elements whether together with the identity they form a subgroup. But that's a lot of tedious work, and therefore it makes sense to save work by classifying the subgroups in a way so that the groups in each class can be handled in an uniform manner. And it turns out that classifying by the number of “basic elements” is a useful criterion for that.



*Why must it “contain $\{b+c,a+b+c\}$”?


Because those are the only two elements that remain after excluding the others.



*I can't grasp what this Possibility 3 is saying. Why "out of luck"? How does $a+b+c$ $\implies$ “any third element will be something like $a+b$ and $(a+b+c)+(a+b)=c$”?


There are three basic elements ($a$, $b$, $c$), three sums of two basic elements ($a+b$, $a+c$, $b+c$), and one sum of all three elements ($a+b+c$). We are considering the case that no basic element is in the set. Now if your group contains $a+b+c$, then since you can't take the second element from the basic elements, you have no choice but take a sum of two elements, for example $a+b$. But as mentioned before, once you have two elements of order $2$, the third one in your subgroup is the sum of those two. Which in this case is $(a+b)+(a+b+c)$. But since we are in an abelian group, and $a$ and $b$ are elements of order $2$, this sum evaluates to $c$. But $c$ is a basic element, in contradiction to the assumption that no basic element is in the group.



*Why can we "ignore the elements of order 4"?


Because the Klein $4$-group does not contain elements of order $4$. Thus any subgroup containing them clearly won't be isomorphic to the Klein $4$-group.



*This isn't obvious to me. How are all Klein groups subgroups of $H=\langle a,b,c\rangle$?


All elements of the product group are made of one element each of the factors. Now the order of such an element of the product is the least common multiple of the individual group element orders.
Now all non-neutral elements of the Klein $4$-group are of order $2$. The only way that the least common multiple is $2$ is if all the factors have order either $1$ or $2$, and at least one of them has order $2$. Note that the only element of order $1$ is the neutral element, that is $0$.
Now the set of tuples $(x,y,z)$ where all of $x$, $y$ and $z$ are of order $1$ or $2$ is exactly the subgroup $H$ mentioned above. Note that $(x,y,z) = (x,0,0) + (0,y,0) + (0,0,z)$, therefore $H$ is generated by non-identity terms of the form $(x,0,0)$, $(0,y,0)$ and $(0,0,z)$ where $x$, $y$ and $z$ are of order $2$. But those are exactly our elements $a$, $b$, $c$.
