Normal subgroup of Engel group

The Engel algebra $\mathfrak g$ is the Lie algebra generated, as a vector space, by four vectors $X_1,X_2,X_3,X_4$ with the only non trivial commutation relations:$$[X_1,X_2]=X_3, \quad [X_1,X_3]=X_4.$$ The Engel group $\mathbb G$ is the only connected and simply connected Lie group whose Lie algebra is $\mathfrak g$.

Now, we denote with $exp\colon \mathfrak g\to \mathbb G$ the usual exponential map and let $exp(\mathbb R X_2):=\{exp(tX_2):t\in \mathbb R\}.$ $exp(\mathbb RX_2)$ is a subgroup of $\mathbb G$. I was wondering if it is also normal or not.

By proposition 9.2 here, if $exp(\mathbb R X_2)$ is normal, then $[\mathfrak g,span\{X_2\}]\subseteq span\{X_2\}$. But, for example, $[X_1,X_2]=X_3\notin span\{X_2\},$ so $exp(\mathbb RX_2)$ is not normal.

Is it correct? Now,is it true that $X_3\notin span\{X_2\}$? I was thinking that, if $X_3\in span\{X_2\},$ then $X_3=aX_2$ for some $a \in \mathbb R$, then $$X_4=[X_1,X_3]=a[X_1,X_2]=aX_3,$$ thus the Engel algebra would be three dimensional, so $X_3\notin span\{X_2\}.$

Is it right?

• Looks fine to me. Although your notes do not mention Engel algebra and the Wikipedia article gives a more general definition than yours of an Engel algebra (in particular it says that in the finite-dimensional case, an Engel algebra and a nilpotent algebra are the same). Why is your algebra "the" Engel algebra? Did you pick it from some exercise? I am just being curious here. Commented Dec 28, 2015 at 12:00
• I have been imprecise above. I've seen that there are several definitions of Engel group, this is the one that interests me, it comes from this article here arxiv.org/abs/1201.6399. Commented Dec 28, 2015 at 12:09

The so-called "Engel-algebra" of dimension $4$ usually has another name, namely it is the standard graded filiform nilpotent Lie algebra $\mathfrak{f}_4$ of dimension $4$. In general, the Lie algebra $\mathfrak{f}_n$ is defined by a basis $(e_1,\ldots ,e_n)$ and nonzero Lie brackets $$[e_1,e_i]=e_{i+1}, \; 2\le i\le n-1$$ We have $[\mathfrak{f}_n,span \{e_2\}]=span \{e_3\}$, which is not contained in $span\{e_2\}$, simply because $e_3\not\in span \{e_2\}$, because $e_1,\ldots ,e_n$ is a basis of the vector space.