For $G$ a Lie group with bi invariant metric, $\nabla_{X}{Y} = \frac{1}{2}[X,Y]$ for left invariant vector fields $X,Y$. Let $G$ be a Lie group with a bi-invariant metric $\langle, \rangle$. Let $X,Y,Z \in \mathfrak{X}(G)$ be (unit) left invariant vector fields on $G$.
Show that $\nabla_{X}Y=\frac{1}{2}[X,Y]$. 
There are three parts to this problem and I don't think the first part uses the fact that these are specifically unit vector fields. 
My attempt: 
I know from a previous exercise, that under the conditions of the problem $\nabla_{X}{X}=0$ for any left invariant vector field $X$. The hint for the problem also says to use the symmetry of the Levi-Cevita connection to prove this. What I tried is as follows: Using symmetry of the connection and properties of $\nabla$,
$\nabla_{X+Y}X-\nabla_{X}(X+Y)=[X+Y,X]$
$\implies \nabla_{X}X+\nabla_{Y}X-\nabla_{X}X-\nabla_{X}Y=[X,X]+[Y,X]$
$\implies \nabla_{Y}X-\nabla_{X}Y=[Y,X]$
Which seems to be useless because it's just the statement of linearity of the connection.

From here I'm stuck.


Since the metric is bi-invariant, $\nabla_ZZ=0$ for all $Z$ as you said. So we have
$$
0=∇_{X+Y}(X+Y)=∇_XX+∇_XY +∇_YX+∇_YY =∇_XY +∇_YX
=∇_YX−∇_XY +2∇_XY = [Y, X] + 2∇_X Y.$$
Therefore, $∇_XY =−\frac{1}{2}[Y,X]=\frac{1}{2}[X,Y]$.
