A simple question from isomorphism types of Lie algebras from root systems Let $L,L'$ be simple Lie algebras over $F$, with maximal toral subalgebras $H,H'$ and corresponding root systems $\Phi,\Phi'$. Suppose $\alpha\mapsto \alpha'$ be an isomorphism between root systems of $L,L'$. 
Write decompositions 
$$L=H\oplus (L_{\alpha} \oplus L_{-\alpha}) \oplus (L_{\beta} \oplus L_{-\beta}) \cdots$$
$$ L'=H'\oplus (L'_{\alpha'} \oplus L_{-\alpha'}') \oplus (L_{\beta'}' \oplus L_{-\beta'}') \cdots $$
With this general set-up, I describe some points of Humphreys Lie algebra, which I don't understand. 
(1) Let $\Delta$ be fundamental root system in $\Phi$ and let $\Delta'$ be the corresponding in $\Phi'$ under isomorphism $\Phi\rightarrow \Phi'$. 
(2) Each $x_{\alpha}$ ($\alpha\in\Delta$) determines unique $y_{\alpha}\in L_{-\alpha}$ s.t. $[x_{\alpha},y_{\alpha}]=h_{\alpha}$ and similarly in $L'$.
(3) Let $D$ be subalgebra of $L\oplus L'$ generated by $(x_{\alpha},x_{\alpha'}')$, $(y_{\alpha},y_{\alpha'}')$, $(h_{\alpha},h_{\alpha'}')$ (for $\alpha\in\Delta, \alpha'\in\Delta'$).
(4) Conceivably, $D$ might contain elements such as $(x_{\alpha},x_{\alpha'}')$ and $(x_{\alpha},2x_{\alpha'}')$, where $x_{\alpha}\in L_{\alpha}$, $x_{\alpha'}'\in L'_{\alpha'}$ for some roots $\alpha,\alpha'$ in which case $D$ would contain all of $L'$, then all of $L$, hence all of $L\oplus L'$.
Here, if $D$ contains $(x_{\alpha},x_{\alpha'}')$ and $(x_{\alpha},2x_{\alpha'}')$ then we see (by algebraic operations) that $D$ contains $(x_{\alpha},0)$ and $(0,x_{\alpha'}')$; so $D$ contains $L_{\alpha}$ and $L_{\alpha'}'$, and so contains $L_{\alpha}\oplus L_{\alpha'}'$; but how it was asserted that $D$ contains $L'$ and $L$?
 A: This is sketch since the complete arguments of mine could be lengthy; I don't have yet reduced it into precise, neat short proof.
As it is noted that in (4), if $(x_{\alpha},x'_{\alpha'})$ and $(x_{\alpha},2x'_{\alpha'})$ for some specific $\alpha\in \Delta$, then 
$$(x_{\alpha},0), (0,x'_{\alpha'})\in D.$$
We claim that $(x_{\beta},0)$ and $(0,x'_{\beta'})$ are also in $D$ for all $\beta\in \Delta$, hence by Prop. 14.2 in book of Humphreys, $D$ should be $L\oplus L'$. 
The crucial point is that Dynkin diagram of (root system of) $L$ is connected; so we can move from $\alpha$ to any other simple root $\beta$ in $\Delta$ by a path, i.e. in other words, successively step-by-step in sense 
$$\alpha \rightarrow \alpha_1 \rightarrow \cdots \rightarrow \beta.$$
Think of first step: from $\alpha$ to $\alpha_1$. There is an edge in the Dynkin diagram from $\alpha$ to $\alpha_1$ means $(\alpha,\alpha_1)$ is non-zero, hence either $\alpha_1-\alpha$ or $\alpha + \alpha_1$ is a root. Suppose $\alpha_1-\alpha$ is a root. Then 
$$[L_{\alpha}, L_{\alpha_1-\alpha}]=L_{\alpha+(\alpha_1-\alpha)}=L_{\alpha_1};$$
interpret this as We can pass from $L_{\alpha}$ to $L_{\alpha_1}$ by applying $ad_{\alpha_1-\alpha}$; so by this process, we can move $L_{\alpha}$ into $L_{\beta}$ by successive iterations by $ad$; so we can cover all $L_{\alpha_i}$ for simple roots $\alpha_i$, hence $L$ is covered. q.e.d.
A: If $(x_\alpha, 0)$ belongs to $D$, then $D$ contains the ideal of $L$ generated by $x_\alpha$, which equlas to $L$ by the simplicity of $L$. So the claim follows.
