I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.

And there is the following problem.

Prove that there exists a group $G$ with $4$ elements which has two elements $\sigma, \tau$ such that $\sigma^2 = \tau^2 = e$ and $\sigma \tau = \tau \sigma$ and $G = \{e, \sigma, \tau, \sigma \tau \}$, where $e$ is a unit element of $G$.

My 1st solution is the following:

Let $G = \{e, a, b, c\}$.
Let $e e = e, e a = a e = a, e b = b e = b, e c = c e = c, a b = b a = c, b c = c b = a, c a = a c = b, a a = e, b b = e, c c = e.$
Then, $e$ is a unit element of $G$ and $e^{-1} = e, a^{-1} = a, b^{-1} = b, c^{-1} = c.$
If we define, $\sigma := a, \tau := b$, then $\sigma^2 = \tau^2 = e$ and $\sigma \tau = c = \tau \sigma.$

But it is a little troublesome to prove the associative law holds.
So I don't like this solution.

My 2nd solution is the following:

Let $S_n = \{1, 2, \cdots, n\}$ be the symmetric group degree $n$ which is greater than or equal to 4.
Let $\sigma := (1 2), \tau := (3 4), e := \mathrm{id}$.
Let $G := \{e, \sigma, \tau, \sigma \tau \}$.
Then, $\sigma^2 = \tau^2 = e$ and $\sigma \tau = \tau \sigma$.
$G$ is a finite set and obviously $G$ is closed under multiplication on $S_n$.
So $G$ is a subgroup of $S_n$.

I don't think this solution is pure because I used $S_n$.

Please tell me other solutions.

  • 3
    $\begingroup$ Consider $(\mathbb{Z}_2\times\mathbb{Z}_2,+)$. Prove that this is a group. Then notice that this group consists of the elements $e=(0,0)$, $\tau=(1,0)$, $\sigma=(0,1)$ and $\sigma\tau =(1,1)$. $\endgroup$
    – Kal S.
    Jan 10, 2019 at 12:23
  • 2
    $\begingroup$ If what you want is just to prove the existence of $G$, then you can pick the free monoid on two elements $\sigma, \tau$, impose the ideal of relations $\sigma^2=e=\tau^2$, $\sigma\tau=\tau\sigma$, and check that $(\tau\sigma)(\tau\sigma)=e$. This way associativity is assured. $\endgroup$
    – Jose Brox
    Jan 10, 2019 at 12:24
  • $\begingroup$ Thank you very much for very simple perfect answer, Test123. $\endgroup$
    – tchappy ha
    Jan 10, 2019 at 12:26
  • 1
    $\begingroup$ Thank you very much, Jose Brox. Maybe this answer is what I wanted, but I don't know "free monoid". I will study "free monid". Thank you. $\endgroup$
    – tchappy ha
    Jan 10, 2019 at 12:27
  • 1
    $\begingroup$ the four 2x2 diagonal matrices whose two nonzero elements are either +1 or -1 forms such a group under matrix multiplication. The two matrices which have a single 1 and a single -1 play the role of $\tau$ and $\sigma$. $\endgroup$ Jan 10, 2019 at 20:44

3 Answers 3


You've already been given most of the entries for the group's Cayley table, so let's just see if we can build the rest of it.

In particular, let $\rho = \sigma\tau = \tau\sigma$ denote the fourth element of the group. Then, just from what you've been given (and the definition of the unit element $e$) we have:

$$\begin{array}{|c|cccc|} \hline {\times} & e & \sigma & \tau & \rho \\ \hline e & e & \sigma & \tau & \rho \\ \sigma & \sigma & e & \rho \\ \tau & \tau & \rho & e \\ \rho & \rho \\ \hline \end{array}$$

All we need now is to fill in the remaining entries involving $\rho$. Let's start in the corner, with $\rho^2$. Since $\rho = \sigma\tau = \tau\sigma$, we have $\rho^2 = (\sigma\tau)^2 = \sigma\tau\sigma\tau = \sigma\sigma\tau\tau = \sigma^2\tau^2 = e^2 = e$.

What about the rest of the entries? Well, we clearly have $\sigma\rho = \sigma\sigma\tau = \sigma^2\tau = e\tau = \tau$. By similar reasoning, we can also see that $\tau\rho = \tau\tau\sigma = \sigma$, $\rho\sigma = \tau\sigma\sigma = \tau$ and $\rho\tau = \sigma\tau\tau = \sigma$. So the full multiplication table looks like this:

$$\begin{array}{|c|cccc|} \hline {\times} & e & \sigma & \tau & \rho \\ \hline e & e & \sigma & \tau & \rho \\ \sigma & \sigma & e & \rho & \tau \\ \tau & \tau & \rho & e & \sigma \\ \rho & \rho & \tau & \sigma & e \\ \hline \end{array}$$

Now we can finally check that this multiplication table satisfies the group axioms. Specifically, the closure axiom is trivial to check, $e$ is an identity by construction, and we can see that every element is its own inverse. So all that remains to be verified is that the group operation is associative. (We kind of assumed that it was when we filled in the table above, but that just means that if the operator is indeed associative, those are the values it needs to take for the squares we filled in.)

Since we only have four elements, we could just check all the $4^3 = 64$ combinations by brute force. Or we could be slightly clever and skip all the combinations involving the identity element $e$, since clearly $xy = (ex)y = e(xy) = (xe)y = x(ey) = (xy)e = x(ye)$ for all $x$ and $y$, leaving us only $3^3 = 27$ combinations to check.

Or we could be slightly more clever yet, and observe that the Cayley table we've just constructed is highly symmetric, and in particular that, if $x$, $y$ and $z$ are the three distinct non-identity elements of the group in any order, they will always satisfy the equations $x^2 = e$ and $xy = yx = z$, and thus also all of the following: $$ x(xx) = xe = x = ex = (xx)x, \\ x(xy) = xz = y = ey = (xx)y, \\ x(yx) = xz = y = zx = (xy)x, \\ y(xx) = ye = y = zx = (yx)x, \\ x(yz) = xx = e = zz = (xy)z.$$

This is sufficient to show that the operation $\times$ defined by the Cayley table above is indeed associative for all possible combinations of three elements, and thus $G = (\{e, \sigma, \tau, \rho\}, {\times})$ is indeed a group. (From the symmetry of the Cayley table, we can also clearly see that $\times$ is also commutative, making the group abelian.)

Of course, arguably the easiest solution is simply to recognize that the Cayley table we've constructed above matches that of the Klein four-group, and that the putative group we've just constructed above is thus isomorphic to it. Of course, you have to be aware of the existence of the four-group (and, in particular, of the fact that, up to isomorphism, there are only two groups with four elements — the other one being the cyclic group of order 4, which your group obviously isn't) to make use of this particular shortcut.


First of all, I don't see any problems with working inside $S_n$. Here you have another solution:

Consider the additive group $\mathbb{Z}_2\times \mathbb{Z}_2=\{(0,0),(0,1),(1,0),(1,1)\}$. Then you can pick $\sigma:=(0,1)$ and $\tau:=(1,0)$.

  • 1
    $\begingroup$ Thank you very much for very simple perfect answer, Jose Brox. $\endgroup$
    – tchappy ha
    Jan 10, 2019 at 12:25

You might use this group : $(\mathbf{Z}/\mathbf{2Z})^2$ where the law is the addition on each coordinate. This means : $(a,b) + (a',b') = (a+a', b+b')$ for $a,b,a',b' \in \mathbf{Z}/\mathbf{2Z}$

And $\sigma = (1,0)$, $\tau = (0,1)$.

If you are not familiar with $Z/nZ$ you can check the construction here : https://en.wikipedia.org/wiki/Modular_arithmetic#Integers_modulo_n

This enables you to avoid to check the associativity, because it is already known. Also, you dont have to check the commutativity (i.e $\sigma + \tau = \tau + \sigma$) because it is already known to.

In reality, the group mentionned is unique to isomorphism ; it is not hard to check, it might be a nice exercice for you.

  • $\begingroup$ To the proposer: Multiplication modulo $8$ on $G=\{1,3,5,7\}$. $\endgroup$ Jan 10, 2019 at 16:56

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