Why is a vector space necessarily a $G$-module if it is closed under the action of $G$?

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My lecture notes gives the following definition of a $G$-module:

Let $V$ be a vector space over $F$ and let $G$ be a group. Then $V$ is a $G$-module if there is a multiplication of elements of $V$ by elements of $G$ such that for any $u,v\in V$, $a\in F$ and $h,g\in G$ the following holds:

  1. $(au)g=a(ug)$,
  2. $(u+v)g=ug+vg$,
  3. $v(gh)=(vg)h$,
  4. $ve=v$.

Later the notes gives the definition of a $G$-submodule as a subspace $W$ of $V$ which is itself a $G$-module. It states that this definition is the same as saying that $W$ is closed under the action of $G$. Now, I understand that if $W$ is a $G$-module, then it is closed under the action of $G$. But how come the other direction is true? In other words, why is the action of $G$ on $W$ necessarily linear and associative (points 1-3 in the definition of a $G$-module)?

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Let $u,v\in W$, $a\in F$, and $g,h\in G$ as in the definition you provide. Since $W\subset V$, in particular we know that $u,v\in V$, making points 1-4 hold. As Matthew Leingang pointed out in the comments, this relies on the fact that the submodules inherit the structure of their ambient module, allowing us to define all operations regarding $W$ to be the same as those in $V$ restricted to $W$.

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To systematically reiterate @MatthewLeingang's points, and maybe elaborate a little.

When $G$ acts on a vectorspace $V$ (linearly, with associativity $(gh)\cdot v=g\cdot (h\cdot v)$, and $1_G\cdot v=v$, ...), a sub-vectorspace $W$ stable under that action of $G$ (yes, the same action), is a $G$-sub-module.

Yes, a sub-vectorspace $W$ of $V$ might have a different $G$-action, for some external reason... but, still, if it's stable under the original/given action on $V$, it has at least that action...

Further, with that original action, the injection $j:W\to V$ is a $G$-homomorphism, in the sense that $j(g\cdot w)=g\cdot j(w)$, for $w\in W$. This would not be typically so for random other actions of $G$ on $W$.