# Thread: additive and multiplicative groups of a field are not isomorphic

1. ## additive and multiplicative groups of a field are not isomorphic

This is from Dummit and Foote Abstract Algebra.

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

Hint from the book:

Consider three cases: when $|\mathbb{F}|$ is finite, when $-1 \ne 1$ and when $-1=-1$ in $\mathbb{F}$.

The finite part is not bad.

let $|\mathbb{F}|=n$

then $|(\mathbb{F},+)|=n$ but $|(\mathbb{F}, \cdot )|=n-1$

because 0 does not have an inverse in the multiplicative group.

For the next case I think contradiction is the way to go so I supposed that there exists an isomorphism $\phi : (\mathbb{F},\cdot) \to (\mathbb{F},+)$ and that $-1 \ne 1$

I know that $\phi(1)=0$ becuase the identity must map to the identity. Also that

$\phi(1)=\phi[(-1)(-1)]=\phi(-1)+\phi(-1)=0$

and for any $a \in \mathbb{F}$

$\phi[(-1)(a)]=\phi(-1)+\phi(a)$
$\phi(1(-a))=\phi(1)+\phi(-a)$

I know that inverses must get mapped to inverses by an isomorphism.

So I don't see how to get a contradiction So I know I am missing something.

Thanks

TES

2. Originally Posted by TheEmptySet
This is from Dummit and Foote Abstract Algebra.

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

Hint from the book:

Consider three cases: when $|\mathbb{F}|$ is finite, when $-1 \ne 1$ and when $-1=-1$ in $\mathbb{F}$.

The finite part is not bad.

let $|\mathbb{F}|=n$

then $|(\mathbb{F},+)|=n$ but $|(\mathbb{F}, \cdot )|=n-1$

because 0 does not have an inverse in the multiplicative group.

For the next case I think contradiction is the way to go so I supposed that there exists an isomorphism $\phi : (\mathbb{F},\cdot) \to (\mathbb{F},+)$ and that $-1 \ne 1$

I know that $\phi(1)=0$ becuase the identity must map to the identity. Also that

$\phi(1)=\phi[(-1)(-1)]=\phi(-1)+\phi(-1)=0$

and for any $a \in \mathbb{F}$

$\phi[(-1)(a)]=\phi(-1)+\phi(a)$
$\phi(1(-a))=\phi(1)+\phi(-a)$

I know that inverses must get mapped to inverses by an isomorphism.

So I don't see how to get a contradiction So I know I am missing something.

Thanks

TES
suppose there's a group isomorphism $f: (F,+) \longrightarrow F^{\times}.$

1) $\text{char} F = 2$: then $(f(1))^2=f(2)=f(0)=1.$ the equation $x^2=1$ in a field of characteristic 2 has the unique solution x = 1. thus $f(1)=1=f(0),$ which is impossible because $f$ is injective.

2) $\text{char} F \neq 2$: since $f$ is surjective, there exists $a \in F: \ f(a)=-1.$ but then $f(0)=1=(f(a))^2=f(2a).$ thus $2a=0,$ and hence $a=0.$ but then $-1=f(a)=f(0)=1.$ contradiction!