5
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I tried to fill in the steps but I'm still confounded by this solution. $|\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}| = 2\cdot2\cdot4$.

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

First, we can ignore the elements of order 4. $ \color{darkred}{(1.) \text{ Why? This is at the bottom of the link. } } $
You need to simply find two (distinct) elements of order 2. $ \color{darkred}{(2.) \text{ Why simply two distinct elements 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{darkred}{(3.) \text{ How do you flesh these out? Surely not by trial and error? } } $

$ \color{darkred}{(4.) \text{ How do you envisage and envision there are more elements of order 2?} } $
By cause of p. 61 Theorem 6.6, any subgroup of a cyclic group is cyclic.
Hence by reason of my question (4.), the sum of these 2 distinct elements will also have order 2.
Hence $a+b = (1,1,0), \; a+c = (1,0,2), \; b+c = (0,1,2), \; a+b+c = (1,1,2)$.

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{darkred}{(5.) \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. Any two of $\{a,b,c\}$ are in the subgroup.
Then these 2 add to determine the third and there are $\binom{3}{2}$ ways to do this. It induces:

\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. Only one of $\{a,b,c\}$ is in the subgroup.
Pretend we're working with $a$. Then this subgroup can't contain $a+b$ or $a+c$. Why not? If they were, then by group closure, $a + (a + b) = b$ or $a + (a + c) = c$ are in the subgroup.
This contradicts our presupposition only one of $\{a,b,c\}$ is in the subgroup.
Hence it must contain $\{b + c, a + b + c\}$. $ \color{darkred}{(6.) \text{ Why? How did they determine this? } } $ It induces:

$$\begin{aligned} \langle a , a+b+c \rangle &= \{ e , a , b+c , a+b+c \}\\ &= \{ (0,0,0) , (1,0,0) , (0,1,2) , (1,1,2) \} \\ \langle a , a+b+c \rangle &= \{ e , b , a+c , a+b+c \}\\ &= \{ (0,0,0) , (0,1,0) , (1,0,2) , (1,1,2) \} \\ \langle a , a+b+c \rangle &= \{ e , c , a+b , a+b+c \}\\ &= \{ (0,0,0) , (0,0,2) , (1,1,0) , (1,1,2) \} \end{aligned}$$

Possibility 3. 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{darkred}{(7.) \text{ I can't grasp what this is saying. What's going on? } } $

At the bottom of the link, author writes 'All Klein groups are subgroups of this order 2 subgroup of $\mathbb{Z_2} \times \mathbb{Z_4},$ obviously'. $ \color{darkred}{(8.) \text{ I can't see why this is obvious? } } $

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  • $\begingroup$ why $|\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}| = \mathrm{lcm}(2, 2, 4) = 4$? $\endgroup$ – Babak Miraftab Jan 19 '14 at 17:41
  • $\begingroup$ @Babgen: Thanks. I think I fixed it. Is it perfect now? $\endgroup$ – Group Theory Jan 20 '14 at 9:14
  • 1
    $\begingroup$ Order means size when applied to a group, so not yet. What you are looking for is exponent. $\endgroup$ – Tobias Kildetoft Jan 20 '14 at 9:50

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