$\newcommand{\lcm}[0]{\mathrm{lcm}}$I tried to fill in the steps but I'm confounded by this solution. Here $i$ is the identity element, not $e$.

Because $\lcm(2, 2, 2) = 2$ hence all non-identity elements of $\mathbb{Z_2} \times \mathbb{Z_2}\times \mathbb{Z_2}$? have order 2.

Because $|\mathbb{Z_4}| = 4$ hence $\mathbb{Z_2} \times \mathbb{Z_2}\times \mathbb{Z_2}$ has no subgroups $\simeq \mathbb{Z_4} $. $\color{darkred}{(1.) \text{ Is this perfect? }}$
By cause of Fraleigh p. 61 Theorem 6.6, any subgroup of a cyclic group is cyclic.
$\mathbb{Z_2} \times \mathbb{Z_2}\times \mathbb{Z_2}$ is cyclic hence the seven non-identity elements of order 2 generate 7 distinct subgroups. $\color{darkred}{(2.) \text{ Where does the 7 crop up from? Why are these distinct subgroups? }}$

By cause of Theorem 207, $|subgroup|$ divides $|group|$ hence these 7 subgroups are of order $2$.
There are seven subgroups of order 4 that are all isomorphic to the Klein 4-group. $\color{darkred}{(3.) \text{ Where did this spring up from? Why fret about this? }}$
To list these subgroups, number the elements of $\mathbb{Z_2} \times \mathbb{Z_2}\times \mathbb{Z_2}$.
enter image description here

To construct a subgroup of order 4 in $\mathbb{Z_2} \times \mathbb{Z_2}\times \mathbb{Z_2}$, start by choosing any two of the elements of order 2, call them $x,y$ and then add these elements together to get a third element $\color{green}{x + y}$.
$x,y$ has order 2 hence $\color{green}{x + y}$ has order 2. $\color{darkred}{(4.) \text{ Why? Is this true in general if $|x| \neq |y|$? }}$ Notice $y + \color{green}{(x + y)} = x + 2y = x + id \tag{$\dagger$}$ $x + \color{green}{(x + y)} = 2x + y = id + y \tag{$\ddagger$}$ This makes it clear that $\{i, x, y, x+y\}$ is a subgroup of order 4. $\color{darkred}{(5.) \text{ How? I don't see this. }}$

Also notice $\{i, x, y, x+y\}$ could've been constructed by starting from choosing
either $\{y, x + y\}$ or $\{x, x + y\}$. If we chose $\{y, x + y\}$, add these elements to induce $x$ from $(\dagger)$. If we chose $\{x, x + y\}$, add these elements to induce $y$ from $(\ddagger)$.

There are $\dbinom{7 \text{ elements of order 2 } \in \mathbb{Z_2} \times \mathbb{Z_2}\times \mathbb{Z_2} }{2}$ choices to induce any pair of elements of order 2. But we noted in the last paragraph, for each pair of $\{x,y\}$, $\{y, x + y\} and \{x, x + y\}$ are two other pairs that determine the same subgroup of order 4.

Hence conclude there are $21/3 = 7$ distinct subgroups of order 4. They are: $\{i, a, b, a+b\}, \{i, a, c, a + c\}, \{i, a, g,a+g \}, \\ \{i, b, c, b+c\}, \{i, b, e, b+e \}, \qquad \{i, c, d, c+ d\}, \qquad \{i, d, e, d+e\}$.

$\color{darkred}{(6.) \text{ Where did these 7 subgroups crop up from? Are there better ways to find them}\\ \text{than sallying forth one-by-one:}\\ \{i, a, b, a+b\}, ..., \{i, a, f, a+f\},\{i, b, c, b+c\}, ...,\{i, b, f, b+f\},...,\{i, d, e, d+e\},\{i, d, f, d+f\}, \{i, e, f, e+f\} }$

$\color{darkred}{(7.) \text{ How do you know these 7 are subgroups without doing the One-Step Subgroup Test?}}$

Update Jan. 19 2014: I updated questions (2.) and (6.) overhead.


2 Answers 2


(1). I don't know what you mean by perfect, but it is correct. ${\mathbb Z}_4$ has an element of order 4, and ${\mathbb Z}_2^3$ hasn't, so no subgroup of ${\mathbb Z}_2^3$ is isomorphic to ${\mathbb Z}_4$.

(2). ${\mathbb Z}_2^3$ has 8 elements. One of them is the neutral element; all seven others have order 2.

(3). This doesn't have to be immediately clear at this point; the explanation is in the lines below. Up to isomorphism, the only groups of order 4 are ${\mathbb Z}_4$ and ${\mathbb Z}_2^2$, and in view of (1), subgroups of ${\mathbb Z}_2^3$ are not isomorphic to ${\mathbb Z}_4$. Hence the Klein group is mentioned because, whatever subgroup of order 4 you're going to find, it's going to be isomorphic to the Klein group (and not ${\mathbb Z}_4$ as just explained).

(4). This is true in general for an abelian group. The elements $x$ and $y$ are of order 2 and are distinct. Because $x$ is of order 2, $-x = x$, so $y \neq -x$, so $x + y \neq 0$. It does hold that $(x + y) + (x + y) = x + x + y + y = 0$, so $x + y$ has order 2.

(5). $\{i,x,y,x+y\}$ has 4 elements. What is left to show is that it is a subgroup, i.e., that it is closed under the group operations. For inversion: that's easy, as in ${\mathbb Z}_2^3$ every element is its own inverse anyway. For closure under addition: there are two remaining cases $x + (x + y)$ and $y + (x + y)$ and they are treated by $(\dagger)$ and $(\ddagger)$.

(6). The reasoning above says that among the 21 ways of writing down $i, x, y, x+y$ (with $x, y$ distinct and not $i$; and $x,y$ considered the same as $y,x$), there are 7 distinct sets $\{i, x, y, x + y\}$. The seven concrete distinct ways come from just starting to listing them and throwing out duplicates. Start with $\{i, a, b, a+b\}$, then $\{i, a, c, a+c\}$. After that $\{i, a, d, a+d\}$ turns out to be the same as $\{i, a, b, a+b\}$ as $a + d = b$, etc.

Edit: naming all the elements is maybe not the easiest way to list the subgroups of order 4, but is probably the most elementary way to find them. Using the symmetry inherent in ${\mathbb Z}_2^3$ the subgroups of order 4 can be described as follows. The subgroups are ${\mathbb Z}_2 \times {\mathbb Z}_2 \times \{0\}$ (plus two more of this form); the subgroup $\langle (1,1) \rangle \times {\mathbb Z}_2$ (plus two more of this form); the final one is the one generated by $(1,1,0)$ and $(0,1,1)$. They can also be seen as lines in the Fano plane. Anyway, for someone doing this particular exercise, simply listing them seems the best way to go about this.

(7). Right before (5) it is proven that they are subgroups.

  • $\begingroup$ Thanks a lot. Upvoted. I updated questions (2.) and (6.) overhead. Can you please get back to me in your answer? $\endgroup$
    – user53259
    Jan 19, 2014 at 12:25
  • $\begingroup$ For (2), for an element $x$ of order 2, the subgroup generated by $x$ is $\{i,x\}$. So if you have another $y$ of order 2, $\{i,x\}$ and $\{i,y\}$ are distinct as well. For (6), see the answer. $\endgroup$ Jan 19, 2014 at 12:30
  • $\begingroup$ Thanks again. Can you please flesh out your edit to (6.)? How do you determine all those subgroups based on 'symmetry in ${\mathbb Z}_2^3$? And what are the 'plus two more of this form'? $\endgroup$
    – user53259
    Feb 4, 2014 at 5:45
  • $\begingroup$ For (6), it is probably easier to leave the elements as $(1,0,0)$, $(0,1,0)$, etc.; then it's easier to see if you've already seen that group. (Start with $\{(0,0,0),(1,0,0),(0,1,0),(1,1,0)\}$, then $\{(0,0,0),(1,0,0),(0,0,1),(1,0,1)\}$. After that $\{(0,0,0),(1,0,0),(1,1,0),(0,1,0)\}$ we've already had, etc.). As for 'plus two more of this form' in the final part: I meant $\{0\} \times {\mathbb Z}_2 \times {\mathbb Z}_2$ and ${\mathbb Z}_2 \times \{0\} \times {\mathbb Z}_2$; similarly for the next one, but there one of the two you can't write down so neatly. $\endgroup$ Feb 4, 2014 at 8:15
  • $\begingroup$ Thanks a lot. Please let me mull this over. And can you please move your comments into your answer? $\endgroup$
    – user53259
    Feb 4, 2014 at 10:07

First note that $|\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2| = 8$ so the proper nontrivial subgroups are either order $2$ or $4$.

There are $7$ order $2$ subgroups: $h(1, 0, 0)i = {(0, 0, 0),(1, 0, 0)}, $

$h(0, 1, 0)i = {(0, 0, 0),(0, 1, 0)},$

$h(0, 0, 1)i = {(0, 0, 0),(0, 0, 1)}, $

$h(1, 1, 0)i = {(0, 0, 0),(1, 1, 0)},$

$h(1, 0, 1)i = {(0, 0, 0),(1, 0, 1)}, $

$h(0, 1, 1)i = {(0, 0, 0),(0, 1, 1)},$

$h(1, 1, 1)i = {(0, 0, 0),(1, 1, 1)}$.

There are $7$ order $4$ subgroups:

${(0, 0, 0),(1, 0, 0),(0, 1, 0),(1, 1, 0)}, $

${(0, 0, 0),(1, 0, 0),(0, 0, 1),(1, 0, 1)},$

${(0, 0, 0),(1, 0, 0),(0, 1, 1),(1, 1, 1)}, $

${(0, 0, 0),(1, 1, 0),(0, 0, 1),(1, 1, 1)},$

${(0, 0, 0),(1, 1, 0),(0, 1, 1),(1, 0, 1)}, $

${(0, 0, 0),(1, 1, 1),(0, 1, 0),(1, 0, 1)},$

${(0, 0, 0),(0, 1, 1),(0, 0, 1),(0, 1, 0)}$.

  • $\begingroup$ Welcome to MSE. Please use MathJax to format both, your questions and your answers :) $\endgroup$
    – mrtaurho
    Nov 4, 2018 at 15:29

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