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Endomorphisms of Z/2Z

  1. Mar 30, 2010 #1
    Can anyone tell me how many endomorphisms there are for Z/2Z? I think it is

    0 --> 0 and 1 --> 1

    0 --> 0 and 1 --> 0
  2. jcsd
  3. Mar 30, 2010 #2


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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.
  4. Mar 30, 2010 #3
    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.
  5. Mar 30, 2010 #4


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    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).
  6. Mar 30, 2010 #5
    Dear Landau,

    Thanks for clearing things up for me. However, is two the order of Hom(Z_2, Z_2) or are there more?
  7. Mar 30, 2010 #6


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    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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