Galois fixed part of modules. Why does $|(1-\sigma)A||A^G|=|A|$ hold?

Question

Let $L/K$ be a quadratic extension of number field $K$. Let $\sigma$ be a generator of $G=Gal(L/K)$.

Let $A$ be a finite $Gal(L/K)$ module.

Then,

Why does $|(1-\sigma)A||A^G|=|A|$ hold ?

Here, $|A|$ means order of $A$. I'm having difficulty to relate $A^G$ with $(1-\sigma)A$.I'm even beginning to think that this equation may not hold unless some additional structure is imposed on A.

Any hints or reference is also appreciated. Thank you in advance.

Answer 1

The fixed points $A^G$ are exactly the elements of $A$ that are annihilated by $1-\sigma$ (since they are the elements $a\in A$ such that $\sigma(a)=a$). So, there is a short exact sequence $$0\to A^G\to A\stackrel{1-\sigma}\to (1-\sigma)A\to 0$$ from which your cardinality equation follows.