Answer with step by step explanation:
We are given that a set
Let A={1,2,3,4}
We have to prove that it is a group under modulo 5
![a*_n b}=(a\cdot b)/(5)=remainder](https://img.qammunity.org/2020/formulas/mathematics/college/2u20wgdsytpt6nnv8qc1izzzl9pefrpu6i.png)
Closed property:
for all a,b belongs to A
Associative property:It is satisfied property
for all a,b,c belongs to A
Identity :
Where e is identity of group
![2*_51=2,3*_5 1=3,4*_51=4](https://img.qammunity.org/2020/formulas/mathematics/college/7uya8ko6676ikzoan3c0m3onkqh4q1torw.png)
Hence, identity exist ,e=1
Inverse:
![a*_5a^(-1)=e](https://img.qammunity.org/2020/formulas/mathematics/college/vmswayf6tiu5broy9f7atxd145ht3qedql.png)
![2*_53=1,3*_52=1,4*_4=1,1*_5=1](https://img.qammunity.org/2020/formulas/mathematics/college/nvzn95o6i5himcpv9w9dv2emszqp0eortw.png)
Hence, inverse exist of every element
Given set satisfied all properties of group under multiplication modulo.Therefore, givens set is a group under multiplication modulo.
It is a U(5) because a group under multiplication modulo is called U(n) group U(n)={r, gcd(r,n)=1}
We know that order of group U(n)=
![\phi(n)](https://img.qammunity.org/2020/formulas/mathematics/college/rhrizbusxh8zgkrwtgmsxmkn8l1vg4m3tr.png)
Order of U(5)=4
We know that
isomorphic to
![Z_(p^n-p^(n-1))](https://img.qammunity.org/2020/formulas/mathematics/college/wjpv1ksm27d0qu04izgf8j3mk268du13x7.png)
p=5,n=1
U(5)isomorphic to
![Z_(5-1)=Z_4](https://img.qammunity.org/2020/formulas/mathematics/college/xjl72llfzrq0kgd8mfx5oakr7iaa7w6i8b.png)