Context Let $\mathbb{k}$ be an algebrically closed field. Let us consider the additive group $\mathbb{G}_a \simeq \mathbb{k}$ and the multiplicative group $\mathbb{G}_m \simeq \mathbb{k} \backslash \{0\}$.
I wanted to show that these groups are not isomorphic. Here is my argument:
There is no order $2$ elements in $\mathbb{G}_a$. Indeed, if $x$ was a non-zero element of order $2$ then $2x = 0$ then $x=0$ which is contradictory. However, in the other side $-1$ is of order $2$ in $\mathbb{G}_m$.
Is my argument right?
I think that there may be a problem with caracteristic 2 fields. So I'm looking for a better argument that does not have any caracteristic issues.
Thanks in advance.
K. Y.
Your answer is correct when the characteristic is not $2$.
For an answer that works in any characteristic : show that all elements of $\mathbb{G}_a$ have the same (potentially infinite) order, while $\mathbb{G}_m$ has elements of all finite orders.