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This is false for finite fields as the multiplicative groups of finite fields are cyclic, and different cardinalities yield cyclic groups of different cardinalities. But I'm unsure how to proceed for infinite fields.

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I remember asking a similar question question before. – Prism Mar 11 '14 at 8:26
up vote 13 down vote accepted

The multiplicative groups ${\Bbb F}_3(X)^\times$ and $\Bbb Q^\times$ are both isomorphic to the sum of $C_2$ (corresponding to $\Bbb F_3^\times$ and $\Bbb Z^\times$) with a free abelian group of countable rank (generated by the primes of $\Bbb Z$ and the irreducible polynomials in $\Bbb F_3[X]$ respectively).

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Equally, one may take any two number fields with class number $1$ and roots of unity equal to $\{\pm1\}$. – Lubin Mar 11 '14 at 2:08

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