## 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! $\Box$

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!)