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I'm sure that there are a number of ways of answering this in the affine case; here's one. Let's assume $G$ affine and let's say that $k$ is our algebraically closed field. The following argument may not be the slickest one, but it has the benefit of working in arbitrary characteristic with the enveloping algebra replaced by the hyperalgebra. (In positive characteristic, instead of considering a Lie($G$)-stable vector we'd want to consider a hyperalgebra-stable vector. It is not true in positive characteristic that a Lie($G$)-stable vector is $G$-stable).

Now, a $G$-module structure on $M$ is given by a comodule morphism $$c : M \to k[G] \otimes M .$$ Let $U(G)$ denote the enveloping algebra of Lie($G$); equivalently, this is the so-called hyperalgebra of $G$ (hyperalgebra = enveloping algebra when char$(k) = 0$, but not in positive characteristic). We will view $U(G)$ as a subspace of the full linear dual of $k[G]$; namely, $U(G)$ is the subspace of $k[G]^*$ consisting of elements that vanish on some power of the ideal defining the identity. (You can look in, say, Jantzen's book Representations of Algebraic Groups for more details). Note that $v \in M$ is $G$-stable if and only if $c(v) = 1 \otimes v$.

Now, the action of $U(G)$ on $M$ also comes from the comorphism $c$. Namely, for $v \in M$, if $c(v) = \sum f_i \otimes v_i$ then for $X \in U(G)$ we have $$X.v = \sum X(f_i) \cdot v_i ,$$ where $X(f_i)$ is the dual action of $X$ on $f_i \in k[G]$. Let $U(G)^+$ denote the augmentation ideal; this is the two-sided ideal of $U(G)$ generated by the Lie algebra inside of $U(G)$. (One has to change this statement slightly in positive characteristic). Then $v \in M$ is Lie($G$)-stable killed by Lie($G$) if and only if $X.v = 0$ for all $X \in U(G)^+$. (This part only works in characteristic 0; there are slight modifications to be made in positive characteristic).

So now let's assume that $v \in M$ is $U(G)$-stable. killed by Lie$(G)$. Let's write $$c(v) = \sum f_i \otimes v_i .$$ Without loss of generality we may assume that the $v_i$ are $k$-linearly independent. Since $X.v = 0$ for all $X \in U(G)^+$ this implies $$\sum X(f_i) \cdot v_i = 0$$ for all $X \in U(G)^+$. Since the $v_i$ are linearly independent, this means that $X(f_i) = 0$ for all $i$ and for all $X \in U(G)^+$. Now, any element $f \in k[G]$ that is killed by all elements of $U(G)^+$ must be constant. Hence we have $$c(v) = \sum 1 \otimes v_i .$$ But one of the properties of the comodule morphism is that $(\epsilon \otimes Id_M) \circ c = 1 \otimes Id_M$, where $\epsilon$ is the augmentation of $k[G]$. Hence $$\sum 1 \otimes v_i = 1 \otimes v$$ and $v$ is $G$-stable.

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I'm sure that there are a number of ways of answering this in the affine case; here's one. Let's assume $G$ affine and let's say that $k$ is our algebraically closed field. The following argument may not be the slickest one, but it has the benefit of working in arbitrary characteristic with the enveloping algebra replaced by the hyperalgebra. (In positive characteristic, instead of considering a Lie($G$)-stable vector we'd want to consider a hyperalgebra-stable vector. It is not true in positive characteristic that a Lie($G$)-stable vector is $G$-stable).

Now, a $G$-module structure on $M$ is given by a comodule morphism $$c : M \to k[G] \otimes M .$$ Let $U(G)$ denote the enveloping algebra of Lie($G$); equivalently, this is the so-called hyperalgebra of $G$ (hyperalgebra = enveloping algebra when char$(k) = 0$, but not in positive characteristic). We will view $U(G)$ as a subspace of the full linear dual of $k[G]$; namely, $U(G)$ is the subspace of $k[G]^*$ consisting of elements that vanish on some power of the ideal defining the identity. (You can look in, say, Jantzen's book Representations of Algebraic Groups for more details). So, what we need to see is that an enveloping-algebra-fixed vector is also a $G$-stable vector. Note that $v \in M$ is $G$-stable if and only if $c(v) = 1 \otimes v$.

Now, the action of $U(G)$ on $M$ also comes from the comorphism $c$. Namely, for $v \in M$, if $c(v) = \sum f_i \otimes v_i$ then for $X \in U(G)$ we have $$X.v = \sum X(f_i) \cdot v_i ,$$ where $X(f_i)$ is the dual action of $X$ on $f_i \in k[G]$. Let $U(G)^+$ denote the augmentation ideal; this is the two-sided ideal of $U(G)$ generated by the Lie algebra inside of $U(G)$. (One has to change this statement slightly in positive characteristic). Then $v \in M$ is Lie($G$)-stable if and only if $X.v = 0$ for all $X \in U(G)^+$.

So now let's assume that $v \in M$ is $U(G)$-stable. Let's write $$c(v) = \sum f_i \otimes v_i .$$ Without loss of generality we may assume that the $v_i$ are $k$-linearly independent. Since $X.v = 0$ for all $X \in U(G)^+$ this implies $$\sum X(f_i) \cdot v_i = 0$$ for all $X \in U(G)^+$. Since the $v_i$ are linearly independent, this means that $X(f_i) = 0$ for all $i$ and for all $X \in U(G)^+$. Now, any element $f \in k[G]$ that is killed by all elements of $U(G)^+$ must be constant. Hence we have $$c(v) = \sum 1 \otimes v_i .$$ But one of the properties of the comodule morphism is that $(\epsilon \otimes Id_M) \circ c = 1 \otimes Id_M$, where $\epsilon$ is the augmentation of $k[G]$. Hence $$\sum 1 \otimes v_i = 1 \otimes v$$ and $v$ is $G$-stable.

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I'm sure that there are a number of ways of answering this in the affine case; here's one. Let's assume $G$ affine and let's say that $k$ is our algebraically closed field. The following argument may not be the slickest one, but it has the benefit of working in arbitrary characteristic with the enveloping algebra replaced by the hyperalgebra. (In positive characteristic, instead of considering a Lie($G$)-stable vector we'd want to consider a hyperalgebra-stable vector. It is not true in positive characteristic that a Lie($G$)-stable vector is $G$-stable).

Now, a $G$-module structure on $M$ is given by a comodule morphism $$c : M \to k[G] \otimes M .$$ Let $U(G)$ denote the enveloping algebra of Lie($G$); equivalently, this is the so-called hyperalgebra of $G$ (hyperalgebra = enveloping algebra when char$(k) = 0$, but not in positive characteristic). We will view $U(G)$ as a subspace of the full linear dual of $k[G]$; namely, $U(G)$ is the subspace of $k[G]^*$ consisting of elements that vanish on some power of the ideal defining the identity. (You can look in, say, Jantzen's book Representations of Algebraic Groups for more details). So, what we need to see is that an enveloping-algebra-fixed vector is also a $G$-stable vector. Note that $v \in M$ is $G$-stable if and only if $c(v) = 1 \otimes v$.

Now, the action of $U(G)$ on $M$ also comes from the comorphism $c$. Namely, for $v \in M$, if $c(v) = \sum f_i \otimes v_i$ then for $X \in U(G)$ we have $$X.v = \sum X(f_i) \cdot v_i ,$$ where $X(f_i)$ is the dual action of $X$ on $f_i \in k[G]$. Let $U(G)^+$ denote the augmentation ideal; this is the two-sided ideal of $U(G)$ generated by the Lie algebra inside of $U(G)$. (One has to change this statement slightly in positive characteristic). Then $v \in M$ is Lie($G$)-stable if and only if $X.v = 0$ for all $X \in U(G)^+$.

So now let's assume that $v \in M$ is $U(G)$-stable. Let's write $$c(v) = \sum f_i \otimes v_i .$$ Without loss of generality we may assume that the $v_i$ are $k$-linearly independent. Since $X.v = 0$ for all $X \in U(G)^+$ this implies $$\sum X(f_i) \cdot v_i = 0$$ for all $X \in U(G)^+$. Since the $v_i$ are linearly independent, this means that $X(f_i) = 0$ for all $i$ and for all $X \in U(G)^+$. Now, any element $f \in k[G]$ that is killed by all elements of $U(G)^+$ must be constant. Hence we have $$c(v) = \sum 1 \otimes v_i .$$ But one of the properties of the comodule morphism is that $(\epsilon \otimes Id_M) \circ c = 1 \otimes Id_M$, where $\epsilon$ is the augmentation of $k[G]$. Hence $$\sum 1 \otimes v_i = 1 \otimes v$$ and $v$ is $G$-stable.