A group with 4 elements 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.

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. 
