In a bialgebra $(H,m,u,\Delta,\epsilon)$, subspace of primitive elements are $P(H)=\{x\in H:\Delta (x)=x\otimes 1+1\otimes x\}$. I know that if $x,y$ are primitive, then $[x,y]=xy-yx$ is also primitive, so $P(H)$ is a lie subalgebra under the bracket. I also know if $x_1,\cdots, x_n$ are primitive, then $$\Delta(x_1\cdots x_n)=1\otimes x_1\cdots x_n+x_1\cdots x_n\otimes 1+\sum_{\sigma }x_{\sigma(1)}\cdots x_{\sigma(p)}\otimes x_{\sigma(p+1)} \cdots x_{\sigma(n)} $$ where $\sigma$ runs over all shuffles in $S_n$.
In tensor bialgebra $TV$, all elements in $V$ are primitive, so $P(TV)$ contains at least the lie subalgebra $L(V)$ generated by $V$. But how to show that $P(TV)=L(V)$?
I tried to compute $\Delta(\sum a^Ix_I)-a^I(1\otimes x_I+x_I\otimes 1)=0$ ( $I$ is multi-index here) but it turns out a total mess. Besides, I do not know how to present the element in $L(V)$. Did I miss something here? Is there a slick way to do this?
[add] I noticed that $T(V)$ is the universal enveloping algebra of free Lie algebra $L(V)$ by this answer, also here it says that primitive elements are precisely the lie algerba. But these arguments involve PBW theorem. Can we prove this case without it?