TITLE: why are subextensions of Galois extensions also Galois?
QUESTION [10 upvotes]: An algebraic extension of fields $L|K$ is defined to be a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$. 
Apparently in the sequence of field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This is asserted to be an obvious deduction in section 3.2.1 of Robalo Delgados thesis on Galois Categories referenced in nLab-Grothendiecks Galois Theory, I don't see the obviousness...can someone clarify.
What interests me in this characterisation is that it shows one direction of the Galois correspondance, (the other being shown by Artins Lemma) without going through characterising Galois extensions as algebraic, seperable & normal extensions. In fact Delgado shows that this characterisation follows from his definition.
Edit: I've also asked this question on math.overflow as I thought most people misunderstood the question. I'd welcome attempts to formulate it in clearer terms.

REPLY [2 votes]: @MoziburUllah  I am not going to answer the OP's question directly as it is a psychological rather than a mathematical question.  I will answer a very closely related question which I believe is what the OP wishes to know, and let others answer the original question as they see fit. 
The Fundamental Theorem of Galois Theory says that if $L\supseteq K$ is a normal and separable extension of fields, and $G$ the group of all $K$-automorphisms of of $L$, then there is a $1-1$ Galois correspondence taking subgroups $H$ of $G$ to their fixed fields and fields $M$ with $L\supseteq M\supseteq K$ to the group of $M$-automorphisms of $L$. Our proposition is the converse of this Fundamental Theorem. 
Proposition. Let $L\supseteq K$ be an algebraic extension of $K$, and assume there is a $1-1$ Galois correspondence taking subgroups $H$ of $G$ to their fixed fields and fields $M$ with $L\supseteq M\supseteq K$ to the group of $M$-automorphisms of $L$.  Then $L\supset K$ must be separable and normal.
Proof. If $L\supseteq K$ is not separable then some irreducible polynomial must have a multiple root, and this can only happen if 
 ${\mathrm{char}}(K)=p<\infty$ and some nonzero $\alpha\in L\setminus K$ has $\alpha^{p^k}$ separable over $K$ but 
any automorphism of $L$ that fixes $K[\alpha]$ must fix $\left(\bigcup_n\in p^{-n}K[\alpha]\right)\cap L$. By hypothesis $\left(\bigcup_n\in p^{-n}K[\alpha]\right)\cap L=K[\alpha]$ so $L\supseteq K$ must be separable.
Now let $0\ne\alpha \in L$, and let the minimal polynomial of $\alpha$ over $K$ be $p(X)\in  K[X]$. Let the roots of $p$ in some splitting field of $p$ be $\{\alpha\ , \alpha_1\ ,\ \cdots \ ,\ \alpha_{deg(p)}\ \}$ and by Lagrange 150 years before Galois, $p$ is irreducible and the coefficients of monic irreducible $p$ are the elementary symmetric functions of all the (distinct) roots of $p$  so the product $\prod_{i=1}^k \ (X-\alpha_i)$ which has a root in $L$ cannot have coefficients in $K$ unless $k=deg(p)$ so $L\supseteq K$ is normal.
qed