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
[tex]a\times_n b}=\frac{a\cdot b}{5}=remainder[/tex]
Closed property:[tex]a\times_5 b=\frac{a\times b}{5}=remainder \in A [/tex]for all a,b belongs to A
Associative property:It is satisfied property
[tex]a\times_5(b\times_5c)=(a\times_5b)\times_5c [/tex] for all a,b,c belongs to A
Identity :[tex]a\times_5e=a[/tex] Where e is identity of group
[tex]2\times_51=2,3\times_5 1=3,4\times_51=4[/tex]
Hence, identity exist ,e=1
Inverse:[tex]a\times_5a^{-1}=e[/tex]
[tex]2\times_53=1,3\times_52=1,4\times_4=1,1\times_5=1[/tex]
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)=[tex]\phi(n)[/tex]
Order of U(5)=4
We know that [tex]U(p^n)[/tex] isomorphic to [tex] Z_{p^n-p^{n-1}}[/tex]
p=5,n=1
U(5)isomorphic to [tex] Z_{5-1}=Z_4[/tex]
