Let $G$ be simply connected chevalley group over a field $K$. I am following the notations as in 'Lectures on Chevalley group' by Steinberg (Yale lectures). Let $H$ be the subgroup generated by $\{h_{\alpha}(t), \alpha \in \Phi, t \in K\}$. Here $\Phi$ is the set of roots. Then in page $44$ after lemma $28$ the author says that every $h \in H$ can be written uniquely as $h=\prod_{i=1}^{i=l}h_{\alpha_i}(t_i)$ where $t_i \in K^{\times}$ and $\alpha_i$`s are basis elements of the roots.

The problem is that I do not understand its proof. The author states it as a corollary without any proof. Can someone write a proof for me and explain me why is it true?

Following the notations as in lemma $28$ I know that if $H_{\alpha}=\sum n_iH_i$ then $h_{\alpha}(t)=\prod_{i=1}^{i=l}{h_{\alpha_i}(t)}^{n_i}$. After this I do not see any further. How to use the information that $G$ is simply connected? Are these $n_i$`s equal to $1$ when $G$ is simply connected?

Thanks for your help and your effort for writing and explaining the proof.

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