Does equivalence of Lie algebras imply equivalence of the corresponding Lie groups? Ok so I'm confused about the relation between these two concepts.
If I have a Lie Group $G$ I can associate a Lie algebra $\mathfrak{g}$ by taking its tangent space in the identity, with the appropriate bracket.
Now I have three questions (it's actually one):

*

*Say I have $\mathfrak{g}$ and $\mathfrak{h}$ Lie algebras of two Lie groups $G$ and $H$. Does $\mathfrak{g} \simeq \mathfrak{h}$ imply $G \simeq H$ (as Lie groups, obviously) ? (I think no)


*Now the opposite: if I have $G \simeq H$, can I say $\mathfrak{g} \simeq \mathfrak{h}$? (I think yes)


*What can I say in case one, if the answer is no?
Let me give you an example: talking about Spin groups, Wikipedia states that the accidental isomorphisms between low dimensional Spin groups and classic Lie groups are due to root system isomorphisms. But those are a thing of algebras, I don't understand. Can someone give me a brief overview of those concepts? Thanks
 A: 1) was answered by AustinC and Mnifldz in the comments:  there are many examples showing the answer is "no."
2) This is true.   The key idea of the proof is the following lemma:  Any group homomorphism $f:G\rightarrow H$ induces a map $f_\ast:\mathfrak{g}\rightarrow \mathfrak{h}$ in a functorial fashion.  That is, if the group homomorphism is the identity map, then so is the induced map.  Further, the induced map of a composition is the composition of the induced maps.
Assuming this, suppose $f:G\rightarrow H$ is an isomorphism with inverse $g:H\rightarrow G$.  Then $g\circ f$ is the identity on $G$, so the induced map is the identity on $\mathfrak{g}$.  On the other hand, the induced map is the composition $g_\ast \circ f_\ast$, so $g_\ast \circ f_\ast = Id_{\mathfrak{g}}$.  On the other hand, applying the same argument to $f\circ g$ shows that $f_\ast \circ g_\ast = Id_{\mathfrak{h}}$.  This implies that $f_\ast$ is an isomorphism of Lie algebras with inverse $g_\ast$.
3)  You can say a lot, but not everything.  First, since $\mathfrak{g}$ and $\mathfrak{h}$ are computed using the tangent space as the identity, they are indifferent to components other than the identity component of $H$ and $G$.  (This is how Mnifldz's example arises).  But even if $G$ and $H$ are connected, you can still have $\mathfrak{g}$ and $\mathfrak{h}$ isomorphic without $G$ and $H$ being isomorphic.
You can still save something:  If $G$ and $H$ have isomorphic Lie algebras, then there is another (connected) Lie group $K$ with is simultaneously covers both $G$ and $H$.  So, up to covers, $G$ and $H$ are the same.  (This is how AustinC's example arises:  $\mathbb{R}$ covers $S^1$.)
As a simple corollary to this if $G$ and $H$ are simply connected, then an isomorphism on the algebra level does imply that $G$ and $H$ are isomorphic.
There are a few other instances where you can recover something like this.  For example, if $G$ and $H$ are compact Lie groups which are centerless, then they have isomorphic Lie algebras iff they are isomorphic Lie groups.
