# How both definitions of a lie group are equivalent

I can't proof that these definitions of lie groups are equivalent. Can anyone prove it directly?.

1. A Lie group is a set which carries the algebraic structure of a group and the differentiable structure of a smooth manifold such that the mapping $$G\times G \rightarrow G \hspace{1cm} (a,b) \mapsto ab^{-1}$$ is smooth

2. A Lie group is a set which carries the algebraic structure of a group and the differentiable structure of a smooth manifold such that the mappings $$G\times G \rightarrow G \hspace{1cm} (a,b) \mapsto ab$$ and $$G \rightarrow G \hspace{1cm} g \mapsto g^{-1}$$ are smooth

• Presumably the restriction $1\times G\to G$ (i.e. inversion) must be smooth and hence the composition of inversion with $ab^{-1}$ (i.e. multiplication $ab$) is also smooth.
– anon
Apr 4, 2020 at 0:37
• In fact, it suffices to require smoothness of the multiplication, smoothness of the inversion then follows from the inverse function theorem. Apr 4, 2020 at 8:55

Since $$m:(a,b)\rightarrow ab^{-1}$$ is smooth, and $$i(a)=(e,a)$$ is smooth, $$i_v(a)=m\circ i(a)=a^{-1}$$ is smooth.
$$f:(a,b)\rightarrow (a,i_v(b))=(a,b^{-1})$$ is smooth and $$m\circ f(a,b)=ab$$ is smooth.
1. implies $$a$$,
let $$g(a,b)=ab, m=g\circ (Id_G,i_v)$$.