Group Theory – Finding Subgroups of Z2 x Z2 x Z4 Isomorphic to Klein 4-Group

cyclic-groupsgroup-theorysolution-verification

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.

  1. $\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.

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

  2. $ \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.

  1. $ \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 $\{b + c, a + b + c\}$.

  1. $ \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 $(a+b+c)+(a+b)=c.$

  1. $ \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:

  1. There are 7 elements of order 2.
  2. A subgroup contains exactly 3 of them.
  3. 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:

  1. There are 7 points.
  2. Any line contains exactly 3 points.
  3. 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:

http://math.colgate.edu/~kellen/interspace/Fano-V4...

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.

  1. $ \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.

  1. $\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…

Best Answer

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}

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