How Many Endomorphisms Exist for Z/2Z?

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Discussion Overview

The discussion revolves around the number of endomorphisms for the group Z/2Z. Participants explore the definitions and implications of group homomorphisms and ring homomorphisms in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests there are two endomorphisms: one mapping 0 to 0 and 1 to 1, and another mapping 0 to 0 and 1 to 0.
  • Another participant questions the type of algebraic structure being considered, noting that ring homomorphisms may have additional requirements, such as preserving the multiplicative identity.
  • A participant clarifies that they are considering group homomorphisms, which must map identities to identities, but expresses uncertainty about whether this rules out the second proposed homomorphism.
  • Further clarification is provided regarding the distinction between group and ring structures, emphasizing that while both endomorphisms are valid in the group context, the ring context introduces additional complexities.
  • One participant confirms that the order of Hom(Z_2, Z_2) is indeed 2, referencing the total number of set functions and the requirement that f(0)=0, which eliminates certain mappings.

Areas of Agreement / Disagreement

Participants generally agree that there are two endomorphisms for Z/2Z in the context of group homomorphisms, but there is some disagreement regarding the implications of ring homomorphisms and the definitions involved.

Contextual Notes

The discussion highlights the potential confusion between group and ring homomorphisms, particularly regarding the preservation of identities and the implications for endomorphisms.

jakelyon
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Can anyone tell me how many endomorphisms there are for Z/2Z? I think it is
two:

0 --> 0 and 1 --> 1

0 --> 0 and 1 --> 0
 
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What kind of algebraic structure do you have in mind? Group? Ring?
Sometimes a ring homomorphism (so in particular an endomorphism) is required to preserve the 1-element, which would rule out the second one.
 
I am thinking of group homomorphisms, so I know that it must map identities to identities. But I didn't think that it could only map identities to identities, thus not ruling out the second homomorphism. But then again, I am not 100% sure this is correct.
 
jakelyon said:
I am thinking of group homomorphisms, so I know that it must map identities to identities. But I didn't think that it could only map identities to identities, thus not ruling out the second homomorphism.
A ring has two operations, addition and multiplication. So there are also two "identities": 0 is the neutral element with respect to addition, 1 is the neutral element with respect to multiplication. These are distinct, unless for the trivial ring. A ring homomorphism is a group homomorphism, and at the same time respects multiplication.

But since we're only talking about the group structure, the above is not relevant. They are both endomorphisms. In fact, more generally, these maps are always endomorphisms for any group: the identity, which sends everything to itself, and the trivial one, which sends everything to the identity element (here 0).
 
Dear Landau,

Thanks for clearing things up for me. However, is two the order of Hom(Z_2, Z_2) or are there more?
 
No, the order is 2. Indeed, there are only four (=2^2) set functions between {0,1} and {0,1}, namely:
0->0, 1->0
0->0, 1->1
0->1, 1->0
0->1, 1->1

The requirement that f(0)=0 rules out the last two.
 

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