Let f:(Z4,+) --> (Z8,+) be a homomorphism such that f(1) = 6. Find f(3) ?

f(3) = f(1+1+1) = f(1)+f(1)+f(1) = 6+6+6 = 2

lime wrote:f(3) = f(1+1+1) = f(1)+f(1)+f(1) = 6+6+6 = 2

Thanks lime.

Can you explain how you reached from 6+6+6 = 2 ?

I might be missing some basic fundamentals here.

Thanks

Since 6 + 6 + 6 = 18, 18 in Z8 is 18 mod 8 = 2.

congvan wrote:Since 6 + 6 + 6 = 18, 18 in Z8 is 18 mod 8 = 2.

Thanks congvan.

why did you use the modulus operation ?

Since Z8 only contains the integer mod 8, i.e. {0,1,2,3,4,5,6,7}. Generally Zn only contains the integer mod n.

congvan wrote:Since Z8 only contains the integer mod 8, i.e. {0,1,2,3,4,5,6,7}. Generally Zn only contains the integer mod n.

So any Zn Group with '+' as binary operation contains following values :

Int mod n ?

Secondly what about Zn Group with 'x' as binary operation ?

mathQ wrote:So any Zn Group with '+' as binary operation contains following values :

Int mod n ?

Secondly what about Zn Group with 'x' as binary operation ?

Yes. Its the cyclic group.

Zn with multiplication will only be a group if n is prime, otherwise not all elements will have inverses. If it is a group, then the operation is just taking two numbers and multiplying them mod n just as before.

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