For a finite field $K$ with $char(K) \neq 2$, showing $[K^*:(K^*)^2]=2$ in two different ways.
Let $K$ be a finite field with $char(K) \neq 2$. The mapping $a \rightarrow a^2$ is a homomorphism from $K^*$ to itself with kernel of order $2$ (the kernel will have $1$ and $-1$ only, right???). Hence the image $(K^*)^2$ has order half that of $K^*$, and so $[K^*:(K^*)^2]=2$.
Is this correct? I'm a little iffy about calling the image of the morphism $(K^*)^2$ since I imagine $(K^*)^2=\{ab:a,b \in K\}$ where as i think the image of the morphism as equaling $\{a^2:a \in K\}$.
Also, the fact that $[K^*:(K^*)^2]=2$ is said to follow directly from the fact that $K^*$ is cyclic. I guess if $\left<a\right>=K^*$ then $\left<a^2\right>=\left<a\right>^2=(K^*)^2$ so this makes sense.
Any further insights would be greatly appreciated... Thank you!