# Existence of a group isomorphism between $(\mathbb K,+)$ and $(\mathbb K^\times,\cdot)$

Let $$(\mathbb K,+,\cdot)$$ be a field.

Is there a group isomorphism between $$(\mathbb K,+)$$ and $$(\mathbb K^\times,\cdot)$$ ?

The answer should clearly be negative.

I tried to proceed via contradiction, but this has not led me very far.

Since it's some kind of "trick problem", I'm merely looking for hints.

• But what about a finite field, $\mathbb{F}_{p^n}$? Commented Dec 25, 2014 at 10:17
• If you think the answer is clearly negative, you should be trying to exhibit a counterexample, not trying to use contradiction. Commented Dec 25, 2014 at 10:29
• You haven't specified a quantifier on the field. Is the question whether there ever exists such an isomorphism, or whether there always exists such an isomorphism? If the latter, you just need to exhibit a single field for which such an isomorphism doesn't exist (a finite field will do). Commented Dec 25, 2014 at 10:31
• That still doesn't specify the quantifier. Do you want to know whether there always exists such an isomorphism or whether there ever exists such an isomorphism? Again, the answer to the first question is straightforwardly no, and the proof is you exhibit a single counterexample. Commented Dec 25, 2014 at 10:35
• @MarcGato this is trivially true in a finite field, but wrong if $\mathbb K = \mathbb R$ for example. Commented Dec 25, 2014 at 10:53

If the field $k$ does not have characteristic $2$, then $-1$ is an element of order $2$ in $k^{\times}$. But by hypothesis, since $k$ does not have characteristic $2$, the additive group of $k$ is a vector space over the prime subfield of $k$, which is not $\mathbb{F}_2$, so it has no elements of order $2$. Hence it cannot be isomorphic to the multiplicative group. If $k$ has characteristic $2$, then the additive group of $k$ contains only elements of order $2$. On the other hand, $k$ is infinite, but the equation $x^2 = 1$ admits at most two solutions in a field, so again the additive group cannot be isomorphic to the multiplicative group.