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If $G$ is a finite group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.

Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r>0$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.

Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.

Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.

If $G$ is a finitely generated group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.

Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r>0$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.

Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.

Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.

If $G$ is a finite group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.

Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.

Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.

Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.

If $G$ is a finite group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.

Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r>0$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.

Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.

Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.

If $G$ is a finite group which is torsion free nilpotent of class $n$, the $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.

Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.

Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation pf $G$ induces a representation of $\mathfrak g(k)$.

Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.

If $G$ is a finite group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.

Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.

Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.

Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.