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Is {1, 2, 3,4) a group under .5 (multiplication modulo 5)? If so, is it isomorphic to 24 or to Z2 x Z2?

User Hsgubert
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1 Answer

4 votes

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

Closed property:
a*_5 b=(a* b)/(5)=remainder \in Afor all a,b belongs to A

Associative property:It is satisfied property


a*_5(b*_5c)=(a*_5b)*_5c for all a,b,c belongs to A

Identity :
a*_5e=a Where e is identity of group


2*_51=2,3*_5 1=3,4*_51=4

Hence, identity exist ,e=1

Inverse:
a*_5a^(-1)=e


2*_53=1,3*_52=1,4*_4=1,1*_5=1

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)

Order of U(5)=4

We know that
U(p^n) isomorphic to
Z_(p^n-p^(n-1))

p=5,n=1

U(5)isomorphic to
Z_(5-1)=Z_4

User Radleybobins
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