Proof that every vector field on a Lie group is left-invariant I am just starting a course on Lie groups and I'm having some difficulty understanding some of the ideas to do with vector fields on Lie groups.
Here is something that I have written out, which I know is wrong, but can't understand why:

Let $X$ be any vector field on a Lie group $G$, so that $X\colon C^\infty(G)\to C^\infty(G)$.
  Write $X_x$ to mean the tangent vector
  $X_x\in T_x G$ coming from evaluation at $G$, that is, define
  $X_x(-)=(X(-))(x)$ for some $-\in C^\infty(G)$.
  We also write $L_g$ to mean the left-translation diffeomorphism $x\mapsto gx$.
Now
  \begin{align}
  X_g(-) = (X(-))(g) &= (X(-))(L_g(e))\\
  &= X(-\circ L_g)(e)\\
  &= X_e(-\circ L_g) \\
  &= ((DL_g)_eX_e)(-).
\end{align}
  Using this we can show that $((L_g)_*X)_{L_g(h)}=X_{L_g(H)}$ for all $h\in G$, and thus $(L_g)_*X=X$, i.e. $X$ is left-invariant.

I'm sure that the mistake must be very obvious, but I'm really not very good at this sort of maths, so a gentle nudge to help improve my understanding would be very much appreciated!

Your problem is with the equality $(X(f))(L_g(e))= X(f\circ L_g)(e).$ Note that in $(X(f))(L_g(e))$ you first get the derivative of $f$ with respect to $X$ evaluated at $g.$ But, in  $X(f\circ L_g)(e)$ you modify the function by a left translation. It holds that $(f\circ L_g)(e)=f(g)$ but you cannot say anything at nearby points, which is essential to get $X(f\circ L_g)(e).$ 
