Consider a local (commutative unital) ring $(A,\mathfrak{m})$ and a subgroup $G$ of $1+\mathfrak{m}$. Is it true that $G=1+I$ where $I \subseteq$ is an ideal?
NOTE: in my case I have A artinian and $G$ finite. You can assume it. The statement does not seem true but I can't find a counterexample.
The answer is no. The point is that $G-1$ has to be closed under multiplication of its own elements (at least provided that $G-1 \subseteq \mathfrak{m}$ is a subgroup) but not closed under multiplication by an arbitrary element of the ring. This leads to a natural counterexample: taking $I$ to be a contraction of an ideal to a subring.
Example:
Consider $A=\mathbb{C}[X]/(X^2)$, so that $\mathfrak{m}=(X).$ Then $1+((X)\cap \mathbb{R}[X]/(X^2)) \subseteq 1 + \mathfrak{m}$ is a subgroup, but $(X) \cap \mathbb{R}[X]/(X^2)$ is clearly not an ideal of $A$.
(Remark: Taking an extension of finite fields in the above example instead of $\mathbb{R} \subseteq \mathbb{C}$ should provide an example where $G$ is additionally finite.)