Lie Group has non degenerate two form Show that

Every even dimensional Lie Group has a non degenerate two form.

How does one answer this question?
I can see it's true for $R^{2n}$ and I was thinking about pulling back the forms on $R^{2n}$ on coordinate patches. But I don't know how to show they will agree on intersection (if they will at all).
 A: Since Lie groups act smoothly and transitively on themselves, there's a more elegant approach (which works for constructing all kinds of invariant tensors).
Let $G$, be a Lie group with identity $e$, and $T_eG$ be its Lie algebra. For each $g\in G$, there is a left translation operator $L_g:G\to G$ defined by $L_g(h)=gh$. These are all diffeomorphisms, which we can use to take a tensor defined at a point and transport it to every other point.
Let $F$ be a tensor on $T_eG$ (e.g. a nondegenerate two-form, a inner product, volume form, etc.). We can define an extension $\widetilde{F}$ of $F$ to a tensor field on all of $G$, using the pushforward of left translations:
$$
\widetilde{F}|_g=(L_g)_*F
$$
(It might be worth proving that this field is smooth.) This tensor field is automatically invariant under left translation, and any retains any pointwise properties of $F$ (e.g. (anti)symmetry, nondegeracy, nonvanishing, etc.). It is in fact the unique left-invariant tensor field with $\widetilde{F}|_e=F$. This reduces your problem to finding a nondegerate $2$-form at the identity.
