Group Homomorphisms from Z/<5> to Z/<5>: A Brute Force Approach

[mathematician]

Homework Statement

Find all homomorphisms from Z/<5> into Z/<5>.

The Attempt at a Solution

Is this a brute force question where we consider all the possibilities for the function?
i.e f(0)=0,1,2,3,4
But that would still be combinatorially difficult.

 

[ircdan]

If f(1) =0, then f(x) = 0 for all x. Otherwise, there is k !=0 s.t. f(x) = kx for all x. So there are exactly 5 homos. I'll let you fill in the details(I left out several).

 

[MathematicalPhysicist]

Z/<5>={z<5>|z in Z}
but: <5>={5^0,5^1,...}
from your notation i understnad that <5> should be 5Z.
first you need to specify if it's with respect to addition or multiplication.
let's assume it's addition then 0 should be mapped to zero, and then you need
to map everything that the operation will stay as it is, i mean that f(a+b)=f(a)+f(b)
for example f(4+1)=f(5)=f(0)=0=f(4)+f(1) then the only couple for f(1) or f(4) are 2,3 or 4 and 1 or both zero, and this way you keep mapping between them.
so as you said it's a brute force type of question.

 

[ircdan]

edit

 

[Mystic998]

Presumably, <5> indicates the ideal generated by 5, and not the group.

 

[HallsofIvy]

Presumably, <5> indicates the ideal generated by 5, and not the group.

Yes, and so Z/<5> is the quotient group, the "group of integers modulo 5" as everyone is assuming.

Homework Statement

Find all homomorphisms from Z/<5> into Z/<5>.

The Attempt at a Solution

Is this a brute force question where we consider all the possibilities for the function?
i.e f(0)=0,1,2,3,4
But that would still be combinatorially difficult.

No, not at all. There are only 5 such homomorphisms, not 5!= 120 as you may be thinking; you cannot just assign values arbitrarily. If f is a homomorphism then f(a+ b)= f(a)+ f(b) so f(2)= f(1)+ f(1), f(3)= f(1)+ f(1)+ f(1), and f(4)= f(1)+ f(1)+ f(1)+ f(1). (Of course f(0)= 0 for every homomorphism.) Every thing depends entirely on what f(1) is and there are 5 possible values for that.