# Center of a nilpotent group

I'm trying to solve an exercise from Humphreys' Linear algebraic groups. It asks to show that center Z of a nilpotent algebraic group G has positive dimension if G has positive dimension.

I'm trying to consider the lower central series of G, say $$C_{0}G=G \supset C_{1}G \supset ....C_{n-1}G\supset C_n{G}=\{e\}$$. By definition of being nilpotent, we get that $$C_{n-1}G\subset Z$$ and $$C_{n-1}G\neq \{e\}$$. I'm stuck on how to proceed further.

There are 2 things which might be useful: $$C_{n-1}G$$ is closed and normal in G and if G is connected then the center Z has positive dimension. Any help is appreciated.

Choose the biggest $$i$$, s.t. $$C^{i}G$$ is not discrete, then $$C^{i}(G)^{0}$$ has positive dimension, and $$C^{i+1}(G)$$ is discrete. Now take any element $$g\in G$$, consider the map $$C^{i}(G)^{0}\longrightarrow C^{i+1}(G)$$, $$x\longrightarrow [x,g]$$.
Suppose that $$C_{n-1}G$$ is discrete. Let $$x$$ an element of $$C_{n-2}G$$ and $$G_0$$ the connected component of the identity of $$G$$. Consider the map $$f_x:C_0\rightarrow G$$ defined by $$f_i(g)=[x,g]=xgx^{-1}g^{-1}$$, the image of $$f_x$$ is contained in $$C_{n-1}G$$. Since $$C_{n-1}G$$ is discrete and $$C_i$$ connected, we deduce that $$f_x$$ is constant, and for every $$g\in G, f_x(g)=f_x(x)=1_G$$. This implies that $$C_{n-2}G\subset Z(G)$$.
If $$C_{n-2}G$$ is not discrete. done, if not we repeat the argument with $$C_{n-3}G$$, and we stop for the lowest $$i$$ such that $$C_{n-i}G$$ is not discrete but $$C_{n-i+1}G$$ is discrete. $$C_{n-i}G$$ is in the center of $$G$$.
This shows that the center of $$G_0$$ has dimension $$>0$$.
Consider the map $$g:G\rightarrow Aut(Z(G_0))$$ defined by $$g(x)(y)=xyx^{-1}$$, the kernel of $$g$$ contains $$G_0$$. $$g$$ induces a morphism $$h:G/G_0$$. Remark that $$h$$ induces a morphism $$Lie(h):G/G_0\rightarrow Aut(Lie(Z(G_0))$$ by $$Lie(h)(A)={d\over{dt}}_{t=0}h(exp(tA))$$. The image of $$h$$ is a nilpotent subgroup of the group of automorphism of $$Z(G_0)$$. The image of $$Lie(h):G/G_0\rightarrow Aut(Lie(Z(G))$$ is also nilpotent. We deduce that there exists a non zero vector $$x$$ of $$Lie(Z)$$ such that $$Lie(h)(g)(x)=x$$ for every $$g\in G/G_0$$. $$x$$ is in the center of $$G$$
• Thank you for your answer. I have one question: Its clear that when $C_i$ is the connected component containing $x$, then $f_x=id_G$. How do we show $f_x=id_G$ when the connected component doesn't contain $x$? Jun 4, 2019 at 0:58
• Thank you for your help. I'm confused with the notation. From $x\in Lie(Z)$, how do we get that $x\in Z(G)$? Jun 4, 2019 at 14:40