 Solution

Put $F=L^G$ and $n=|G|$. The automorphism-count bound for a finite field extension givesFor the reverse inequality, let $x_1,\dots ,x_m$ be linearly independent over $F$. We claim that the $n\times m$ matrixhas linearly independent columns over $L$. If not, choose a nonzero relationwith the fewest nonzero coefficients, and normalize one coefficient to $c_1=1$. Applying $h\in G$ and reindexing the rows gives another relation with coefficients $h(c_j)$. Subtracting eliminates the first term, so minimality forces $h(c_j)=c_j$ for every $h$ and $j$. Thus every $c_j\in F$. The row for the identity automorphism then contradicts the $F$-linear independence of the $x_j$.

Therefore $m\le n$. Taking an $F$-basis of $L$ gives $[L:F]\le n$, and henceThis proves the degree assertion in the Artin fixed-field theorem.