How Many Endomorphisms Exist for Z/2Z?

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SUMMARY

The discussion confirms that there are exactly two endomorphisms for the group Z/2Z. These endomorphisms are defined as follows: the identity map (0 → 0 and 1 → 1) and the trivial map (0 → 0 and 1 → 0). The conversation clarifies that while ring homomorphisms must preserve both addition and multiplication identities, the focus here is solely on group homomorphisms, which only require the preservation of the identity element. Thus, the order of Hom(Z_2, Z_2) is definitively two.

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  • Understanding of group theory concepts, specifically endomorphisms.
  • Familiarity with the structure of Z/2Z as a group.
  • Knowledge of homomorphisms, particularly group homomorphisms.
  • Basic understanding of ring theory and the distinction between group and ring homomorphisms.
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  • Study the properties of group homomorphisms in detail.
  • Explore the concept of endomorphisms in various algebraic structures.
  • Learn about the implications of ring homomorphisms and their requirements.
  • Investigate the structure and properties of other finite groups similar to Z/2Z.
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Mathematicians, algebra students, and anyone interested in group theory and its applications, particularly those studying endomorphisms and homomorphisms in algebraic structures.

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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