Show that the set $G=\{f_1, f_2, f_3, f_4, f_5, f_6\}$ where the functions are defined by

$f_1(x)=x \ \ \ \ f_2(x)=1-x \ \ \ \ f_3(x)=x/(x-1) f_4(x)=1/x \ \ \ \ \ f_5(x)=1/(1-x) \ \ \ \ \ f_6(x)=1-(1/x)$

is a group under composition of function. Frame the composition table.

Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 8 Marks

Year: Dec 2015

Let G be the set $\{f_1, f_2, f_3, f_4, f_5, f_6\}$. First we have to check that the composition of any two elements in G is also in G. We build the whole multiplication table:

This shows that composition is well-defined. As we have already remarked, f1 is the left- and right-sided identity for G. Furthermore, we can read from the multiplication table that

so every element has a 2-sided inverse.

Hence set G is a group under composition of function.