# Galois correspondence via group algebras

Suppose $G$ is a finite group of automorphisms of a field $E$. Form the ring $E(G)$ which is the subring of the ring of endomorphisms of $E$ generated by $G$ and multiplication by elements of $E$. Note that if $\lambda\in E$ then the composite $g\cdot \mu_\lambda$ equals the composite $\mu_{g(\lambda)} \cdot g$. Show that the $E$-subalgebras of the group ring $E(G)$ are those of the form $E(H)$ for $H$ a subgroup of $G$.

Note By an $E$-subalgebra we mean those subrings that contain all the multiplications by elements $\lambda \in E$.

1. Consider such a subring $S$, and note that $G\cap S=H$ is a submonoid of $G$, and since $G$ is finite, it is a subgroup. Evidently $E(H)\subseteq S$ by the condition that $S$ be an $E$-subalgebra. Assuming this is a proper inclusion, pick an element of the form $a_1s_1+\cdots+a_rs_r$ such that no $s_i$ lies in $H$ and each $a_j$ is nonzero, with $r$ minimal. Note that $r>1$ since else $a_1s_1\in S\cap G=H$. Now choose $x$ such that $s_1(x)\neq s_2(x)$ and note that $\sum a_i s_i\cdot x=\sum a_i s_i(x) s_i$ lies in $S$, and subtracting $s_1(x)\sum a_is_i$ from this gives a contradiction by the minimality of $r$. This is the same trick one uses to show Dedekind’s theorem that characters of monoid onto fields are linearly independent!