Order of Homomorphisms from Z to Z/2Z - Help Needed

[jakelyon]

I have seen that there exists two group homomorphisms from Z to Z/2Z. However, I cannot seem to understand this. I mean I know that there exists a trivial grp hom. which sends all of Z to 0 in Z/2Z. But I cannot think of anymore. Any help?

 

[Office_Shredder]

Well, Z is generated by 1. There are two places 1 can map to, 0 or 1 (mod 2). If it doesn't map to 0, it must map to 1. Do you see what function that really is?

 

[jakelyon]

You mean that I map all even to 0 and all odd to 1?

 

[Landau]

Yes :)
If f(1)=1 (mod 2), then f(n)=n (mod 2), which is 0 if n is even and 1 if n is odd.

 

[blue_raver22]

I can provide some clarification on the concept of group homomorphisms from Z to Z/2Z. A group homomorphism is a function that preserves the group structure, meaning that the operation between elements in the original group is preserved in the resulting group. In this case, Z and Z/2Z are both additive groups, meaning that the operation between elements is addition.

The trivial group homomorphism you mentioned is the function that maps all elements of Z to the identity element (0) in Z/2Z. This is a valid homomorphism because it preserves the group structure, as any element added to 0 in Z/2Z will still result in 0.

However, there is also another group homomorphism from Z to Z/2Z, which is the function that maps all even numbers in Z to 0 and all odd numbers to 1 in Z/2Z. This is also a valid homomorphism because it preserves the group structure, as any even number added to another even number will result in an even number, which is mapped to 0 in Z/2Z. Similarly, any odd number added to another odd number will result in an even number, which is also mapped to 0 in Z/2Z.

I hope this helps clarify the existence of two group homomorphisms from Z to Z/2Z. It is important to note that there may be other possible homomorphisms, but these two are the most commonly used and easiest to understand.