MHB Isomorphic Lie groups with isomorphic Lie algebras?

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Two simply connected Lie groups, G and H, with isomorphic Lie algebras are shown to be isomorphic through a proof utilizing a key theorem. The proof establishes a Lie algebra isomorphism and constructs corresponding Lie group homomorphisms, demonstrating that the identity maps are preserved under composition. This leads to the conclusion that both homomorphisms are inverses of each other, confirming the isomorphism between G and H. The discussion highlights the relationship between Lie algebras and their corresponding Lie groups, emphasizing the significance of simply connectedness in this context. Ultimately, the problem illustrates a fundamental result in the theory of Lie groups and algebras.
Chris L T521
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Here's this week's problem!

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Problem
: Let $G$ and $H$ be two simply connected Lie groups with isomorphic Lie algebras. Show that $G$ and $H$ are isomorphic.

The following theorem can be used without proof in your solution:

Theorem
: Suppose $G$ and $H$ are Lie groups with $G$ simply connected, and let $\mathfrak{g}$ and $\mathfrak{h}$ be their Lie algebras. For any Lie algebra homomorphism $\varphi:\mathfrak{g}\rightarrow\mathfrak{h}$, there is a unique Lie group homomorphism $\Phi:G\rightarrow H$ such that $\Phi_{\ast} = \varphi$ (where $\Phi_{\ast}$ denotes the pushforward of $\Phi$).

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No one answered this week's problem. You can find the solution below.

[sp]Proof: Let $\mathfrak{g}$ and $\mathfrak{h}$ be the Lie algebras, and let $\varphi:\mathfrak{g}\rightarrow\mathfrak{h}$ be a Lie algebra isomorphism between them. Then by the theorem mentioned, there are Lie Algebra $\Phi:G\rightarrow H$ and $\Psi:H\rightarrow G$ satisfying $\Phi_{\ast}=\varphi$ and $\Psi_{\ast}=\varphi^{-1}$. Both the identity map of $G$ and the composition $\Psi\circ\Phi$ are maps from $G$ to itself whose induced homomorphisms are equal to the identity, so the uniqueness part of the Theorem implies that $\Psi\circ\Phi= \mathrm{Id}_G$. Similarly, $\Phi\circ\Psi = \mathrm{Id}_H$ so $\Phi$ is a Lie group homomorphism.$\hspace{.25in}\blacksquare$[/sp]