Additive and multiplicative groups of a field

Posted: January 26, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

Problem. Prove that the additive group and the multiplicative group of a field are never isomorphic.

Solution. Let F be a field. Let F^{+} and F^{\times} be the additive and the multiplicative group of F respectively. Suppose that F^{+} \cong F^{\times}. Then the number of solutions of the equation 2x=0 in F^{+} must be equal to the number of solutions of x^2=1 in F^{\times}. Now, if \text{char}(F) = 2, then x^2 = 1 has only one solution in F^{\times} but 2x = 0 will have |F| \geq 2 solutions in F^{ + }. If \text{char}(F) \neq 2, then x^2 = 1 has exactly two solutions in F^{\times} but 2x = 0 has only one solution in F^{ + }. Contradiction! \Box

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Comments
  1. mlbaker says:

    if the additive and multiplicative groups of a field are isomorphic then the field is a reductive Lie algebra, because something false implies anything

    • Yaghoub Sharifi says:

      Well, I’m not sure what you mean by that. Any associative commutative algebra (over a field) A is a reductive Lie algebra because by defining [a,b]:=ab-ba=0, a,b \in A, the algebra A becomes an abelian, and hence reductive, Lie algebra.

  2. Chandrasekhar says:

    Hi–

    How do you prove that if the additive and multiplicative groups of a field are isomorphic then F is infinite.

    • Yaghoub Sharifi says:

      Hi
      Well, the reason is clear: two isomorphic groups must have the same cardinality. If F is finite, then the additive group of F will have |F| elements but the multiplicative group of F will have
      |F| – 1 elements and so they can’t be isomorphic. It’s worth mentioning that my proof works perfectly even without this fact.
      (I just removed that part and you should take a look at the proof again!)

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