Show there is no proper subfield that contains $\sigma(E)$ for all $\sigma(E) \in \Gamma(L:K)$ Let $L:K$ be a Galois extension and let E be a subfield of L containing K. Suppose that L is the smallest field containing E and is Galois over K. 

Show there is no proper subfield that contains $\sigma(E)$ for all 
  $\sigma \in \Gamma(L:K)$

My work so far:
Suppose there exists such a proper subfield, $M.$ Clearly $E \subset M$ since, taking $\sigma$ to be the identity, $\sigma(E) = E$. 
We also have that, since $ \sigma(E) \subset M$, $\Gamma(L:M) \subset \Gamma(L:\sigma(E))$
I've tried to prove it by trying to show that the existence of $M$ implies that  $\Gamma(L:K)^f = E$, which contradicts $L:K$ being Galois (unless $E = K$) but I made no progress (as I'm not even sure this is true). And I'm also tried to show that maybe the existence of $M$ implies that $\Gamma(L:K) = \Gamma(L:M)$ which again would contradict $L:K$ being Galois since $M$ is a proper subset but also made no substantial progress.
 A: Hint: if $L/K$ is a finite Galois extension, and $E$ is a subfield of $L$ containing $K$, then the composite field $$\prod\limits_{\sigma \in \operatorname{Gal}(L/K)} \sigma(E) $$ is a Galois extension of $K$.
A: Let $M \subseteq L$ be a field containing $\sigma(E)$ for all $\sigma \in \Gamma = \operatorname{Gal}(L/K)$. We want to show that $L \subseteq M$.
Since $L/K$ is Galois, it is separable, and so $E/K$ is also a separable extension.  So by the primitive element theorem, there exists a $\beta = \beta_1 \in G$ such that $E = K(\beta)$.  Let $f$ be the minimal polynomial of $\beta$ over $K$.  Since $L/K$ is normal, $f$ factors in $L[t]$ as $f(t) = (t - \beta_1) \cdots (t - \beta_n)$.
Now each $K$-isomorphism $K(\beta) \rightarrow K(\beta_i)$, determined by $\beta \mapsto \beta_i$, extends to a $K$-isomorphism $\sigma_i: L \rightarrow L$ (here we are using again that $L/K$ is Galois), and by definition, $\sigma_i$ is an element of $\Gamma$.  Thus $K(\beta_i) = \sigma_i (K(\beta)) = \sigma_i (E) \subseteq M$.  
It follows that $K(\beta_1, ... , \beta_n)$, i.e. the splitting field of $f$ over $K$, is contained in $M$.  The field $K(\beta_1, ... , \beta_n)$ is normal over $K$ (being a splitting field), and separable over $K$ since it is obtained by adjoining separable elements, hence Galois over $K$.  By our hypothesis on $L$, we must have $L = K(\beta_1, ... , \beta_n)$.  Hence $L \subseteq M$.
