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It can be proven that for any vector space $V$ the action of $\mathrm{GL}(V)$ on $V \setminus \{0\}$ is transitive, and its stabilizer is $U^* \rtimes GL(U)$, where $U$ is a complement to the subspace spanned by some non-zero vector from $V$. The proof that I found relies heavily on the fact that any non-zero vector can be extended to a basis, so naturally it's a useless tactic for modules.

Can this proposition be generalized to modules or at least some class of modules larger than vector spaces? What proof strategy would you suggest?

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up vote 6 down vote accepted

Here is one counterexample: for $\mathbb{Z}$-modules, we have that $\text{GL}(\mathbb{Z})=\{\pm1\}$, which does not act transitively on $\mathbb{Z}\setminus\{0\}$.

If we want even the claim that $\text{GL}(M)$ acts transitively on $M\setminus\{0\}$ for any $R$-module $M$ to be true, then it must be true for $R$ treated as an $R$-module, so we need $\text{GL}(R)=R^\times$ to act transitively on $R\setminus\{0\}$, i.e. for any $x,y\in R\setminus\{0\}$ there is a unit $u$ such that $y=ux$. In particular, for every $x\in R\setminus\{0\}$, we have that $x=u\cdot1=u$ for some unit $u$, i.e. every non-zero element is a unit, i.e. $R$ is a field. So this won't be true for anything other than vector spaces.

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