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Let G be a nilpotent Lie group and let $G_0$ be the connected component of e. Assume additionally that $G_0$ is abelian. Consider a map $S\colon G_0\to G_0$ defined by $$g_0 \mapsto \gamma g_0 \gamma^{-1}.$$ Show that it is unipotent map, i.e. there exists $n\in\mathbb{N}$ such that $(S-Id)^n=0$.

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Background, ideas/insights, self work already done...? – DonAntonio Sep 4 '12 at 13:22
Dear Annie, The fact that you write $(S-Id)^n$ suggests to me that you are thinking of $S$ as a linear map, which means that you are probably thinking about $S$ acting on the Lie algebra of $G_0$, rather than $G_0$ itself. (In fact, for a connected nilpotent Lie group, the exponential map induces an isomorphism between the lie algebra of $G_0$ and $G_0$, but it is probably safest to think of them being distinct, or at least to be explicit about it if you intend to identify them.) Regards, – Matt E Sep 4 '12 at 13:47
@BenjaLim: I think a non-trivial nilpotent finite group would be a non-connected nilpotent Lie group of dimension $0$. – Marc van Leeuwen Sep 4 '12 at 14:15
My question comes from a paper which I am trying to read. It seems that it must be something obvious to people who know a little bit about Lie groups since it comes in the middle of a proof of some more involved proposition with no further explanation (the assumption on G_0 being abelian might be redundant for this part of the proof). Since I have little experience with Lie groups, I have no idea how to even start proving such a fact. – Annie Sep 4 '12 at 16:08
@BenjaLim $\gamma$ is a fixed element of G. There are examples of non-connected nilpotent Lie groups: upper-triangular matrices 3x3 with 1's on the diagonal, 0's below the diagonal, an integer entry in the middle row, last column and two more real entries (multiplication is just the standard matrix multiplication). – Annie Sep 4 '12 at 16:10
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