How to prove that a unique inverse exists for every element of this group like structure

Question

Given a quartet $(s,p,m, n)$ and a law

$$(s,p,m,n).(q,r,t,u)=\left(\frac{2}{3}sq, pr,m+(1+t), nu\right)$$

where $s,p,m,n,q,r,t$ and $u$ are in $\mathbb{R} \setminus \{0\}$.

The right and left identities seem to be the same and are $(\frac{3}{2},1,-1,1)$ for the left identity and $(\frac{3}{2},1,-1,1)$ also for the right identity. How can we prove that a unique left and right inverse exists for every element?

Answer 1

What you have is not an operation, it's just a function. Taking $\mathbb{R}^*=\mathbb{R}\setminus\{0\}$, the domain is $(\mathbb{R}^*\times\mathbb{R}^*\times\mathbb{R}^*\times\mathbb{R}^*)^2$ and the image is $\mathbb{R}^*\times\mathbb{R}^*\times\mathbb{R}\times\mathbb{R}^*$ (note the third coordinate). Because we do not have an operation, the terminology of "left inverse" and "right inverse" is not really appropriate.

You can avoid this problem by taking tuples $(a,b,c,d)$ in which $a,b,d\in\mathbb{R}^*$ and $c\in\mathbb{R}$ (so you allow the third coordinate to be $0$, but keep the first, second, and fourth nonzero).

If you do that, then you really have three different, independent operations: two on $\mathbb{R}^*$, one on $\mathbb{R}$, then put together with a direct product.

The first operation on $\mathbb{R}^*$ given by $a\cdot b = ab$, which is well-known to give a group; this is what you do in the second and fourth coordinates. The identity is $1$, the inverse of $a$ is $\frac{1}{a}$.

The second operation on $\mathbb{R}^*$ is $a\cdot b = \frac{2}{3}ab$. This is also a commutative group operation on $\mathbb{R}^*$: the identity is $\frac{3}{2}$; it is associative, since $$\begin{align*} (a\cdot b)\cdot c &= \frac{2}{3}(a\cdot b)c = \frac{4}{9}abc\\ a\cdot(b\cdot c) &= \frac{2}{3}a(b\cdot c) = \frac{4}{9}abc. \end{align*}$$ The inverse of $a$ is the $b$ such that $\frac{2}{3}ab=\frac{3}{2}$, that is, $b=\frac{9}{4a}$, which makes sense because $a\neq 0$.

Finally, the third operation at play occurs in the third coordinate, where we have a map $\mathbb{R}\times\mathbb{R}\to\mathbb{R}$, given by $a\cdot b = a+b+1$. This is also a group: commutative, associative as $(a\cdot b)\cdot c = a+b+c+2 = a\cdot (b\cdot c)$. The identity is $-1$, as $a\cdot -1 = a-1+1=a$; the inverse of $a$ is the unique $b$ such that $a+b+1=-1$, that is $b=-a-2$.

So what we have is three groups, $G_1$, $G_2$, and $G_3$, given as follows:

1. $G_1$ is the group with underlying set $\mathbb{R}\setminus\{0\}$, and operation $a\cdot b = \frac{2}{3}ab$. Identity element is $\frac{3}{2}$, inverse of $a$ is $\frac{9}{4a}$.

2. $G_2$ is the multiplicative group of nonzero reals. The identity is $1$, the inverse of $a$ is $\frac{1}{a}$.

3. $G_3$ is the group with underlying set $\mathbb{R}$ and operation $a\cdot b = a+b+1$. The identity element is $-1$, the inverse of $a$ is $-a-2$.

Your group is the group $G_1\times G_2\times G_3\times G_2$. It is commutative, since each of the factor groups are commutative. The identity element is $(e_1,e_2,e_3,e_2)= (\frac{3}{2},1,-1,1)$; the inverse of $(s,p,m,n)$ is $(\frac{9}{4s},\frac{1}{p},-m-2,\frac{1}{n})$.