To prove that a binary operation on a certain set, say
![](https://latex.codecogs.com/png.latex?G)
makes it a group, we need to show that four properties are satisfied. Let's illustrate this by examples.
Problem 1
Let
![](https://latex.codecogs.com/png.latex?G = \mathbb{Q})
be a set of non-zero rational numbers such that
![](https://latex.codecogs.com/png.latex?a*b = \dfrac{ab}{2})
,
![](https://latex.codecogs.com/png.latex?a, b \in G)
. Show that
![](https://latex.codecogs.com/png.latex?G)
is a group.
(i) Closure: If
![](https://latex.codecogs.com/png.latex?a)
and
![](https://latex.codecogs.com/png.latex?b)
are non-zeros, then
![](https://latex.codecogs.com/png.latex?ab)
is non-zero rational number, and so
(ii) Associative: For
Similarly,
(iii) Identity element: The number two acts as the multiplicative identity, so,
(iv) Inverse: If
![](https://latex.codecogs.com/png.latex?a)
is non-zero, then
![](https://latex.codecogs.com/png.latex?\frac{4}{a})
is a no-zero rational number and serves as the multiplicative inverse of
![](https://latex.codecogs.com/png.latex?a)
, since
Problem 2
Let
![](https://latex.codecogs.com/png.latex?G = \{x\in \mathbb{R}| \ x>1 \})
, i.e., the set of all real numbers greater than
![](https://latex.codecogs.com/png.latex?1)
. Define
![](https://latex.codecogs.com/png.latex?x*y = xy-x-y+2,)
for
![](https://latex.codecogs.com/png.latex?x, \ y \in G.)
Show that the operation
![](https://latex.codecogs.com/png.latex?*)
makes
![](https://latex.codecogs.com/png.latex?G)
a group.
Solution
For
![](https://latex.codecogs.com/png.latex?G)
to be a group, four properties: closure, associative, identity element and inverse must be satisfied.
(i) Closure: If
![](https://latex.codecogs.com/png.latex?a, \ b\in G)
, then
![](https://latex.codecogs.com/png.latex?a>1, \ b>1,)
so
![](https://latex.codecogs.com/png.latex?b-1>0)
and therefore multiplying
![](https://latex.codecogs.com/png.latex?a>1)
by
![](https://latex.codecogs.com/png.latex?b-1>0)
both sides, it follows immediately that
Hence the operation
![](https://latex.codecogs.com/png.latex?*)
is closed on
![](https://latex.codecogs.com/png.latex?G)
.
(ii) Associative: For
![](https://latex.codecogs.com/png.latex?a, \ b, \ c \in G,)
we have
On the other hand, we have
Thus
(iii) Identity:
![](https://latex.codecogs.com/png.latex?2*a = a*2 = 2a - 2 - a + 2 = y.)
Hence, 2 is the identity element for the operation
![](https://latex.codecogs.com/png.latex?*)
.
(iv) Inverse: Given any
![](https://latex.codecogs.com/png.latex?a \in G,)
we need to solve the equation
![](https://latex.codecogs.com/png.latex?a*a^{-1} = 2.)
This gives the equation
This solution belongs to
![](https://latex.codecogs.com/png.latex?G)
since
NB: You can also check that
![](https://latex.codecogs.com/png.latex?a*\left(\frac{a}{a-1}\right) = \frac{a^2}{a-1} - a - \frac{a}{a-1} + 2 = \frac{a^2 - a^2 + a - a}{a-1} + 2 = 2.)