[proofplan]
We identify the tangent space at the identity of the product Lie group $G\times H$ with the direct sum of the tangent spaces at the identities of $G$ and $H$. This gives the underlying vector-space isomorphism $\operatorname{Lie}(G\times H)\cong\mathfrak g\oplus\mathfrak h$. To prove that it preserves brackets, we write the left-invariant vector field on $G\times H$ determined by $(X,Y)$ and observe that it is the product of the left-invariant vector fields determined by $X$ on $G$ and by $Y$ on $H$. The Lie bracket of product vector fields is computed componentwise, which gives the desired formula.
[/proofplan]
[step:Identify the tangent space at the product identity]
Let $e_G\in G$ and $e_H\in H$ denote the identity elements, and let
\begin{align*}
K:=G\times H
\end{align*}
denote the product Lie group with identity
\begin{align*}
e_K:=(e_G,e_H).
\end{align*}
Let
\begin{align*}
\operatorname{pr}_G:K\to G
\end{align*}
and
\begin{align*}
\operatorname{pr}_H:K\to H
\end{align*}
be the smooth coordinate projection maps.
Define a [linear map](/page/Linear%20Map)
\begin{align*}
\Phi:T_{e_K}K\to T_{e_G}G\oplus T_{e_H}H
\end{align*}
by
\begin{align*}
\Phi(Z):=((d\operatorname{pr}_G)_{e_K}(Z),(d\operatorname{pr}_H)_{e_K}(Z)).
\end{align*}
By the definition of the product smooth manifold structure, tangent vectors at $(e_G,e_H)$ are pairs of tangent vectors at $e_G$ and $e_H$, and the map $\Phi$ is the corresponding linear isomorphism.
Since
\begin{align*}
\operatorname{Lie}(G)=T_{e_G}G
\end{align*}
and
\begin{align*}
\operatorname{Lie}(H)=T_{e_H}H,
\end{align*}
the map $\Phi$ gives a vector-space isomorphism
\begin{align*}
\operatorname{Lie}(K)=T_{e_K}K\cong \mathfrak g\oplus\mathfrak h.
\end{align*}
[/step]
[step:Write the left-invariant vector field determined by a product tangent vector]
Let $\mathfrak X(M)$ denote the space of smooth vector fields on a smooth manifold $M$. Let $X\in\mathfrak g$ and $Y\in\mathfrak h$. Let $\widetilde X\in\mathfrak X(G)$ be the left-invariant vector field on $G$ determined by $X$, and let $\widetilde Y\in\mathfrak X(H)$ be the left-invariant vector field on $H$ determined by $Y$. Thus, for each $g\in G$ and $h\in H$,
\begin{align*}
\widetilde X_g=(dL^G_g)_{e_G}(X)
\end{align*}
and
\begin{align*}
\widetilde Y_h=(dL^H_h)_{e_H}(Y),
\end{align*}
where $L^G_g:G\to G$ is the left translation map $a\mapsto ga$, and $L^H_h:H\to H$ is the left translation map $b\mapsto hb$.
Let $\widetilde{(X,Y)}\in\mathfrak X(K)$ be the left-invariant vector field on $K$ determined by $(X,Y)\in T_{e_K}K$. For $(g,h)\in K$, the left translation map
\begin{align*}
L^K_{(g,h)}:K\to K
\end{align*}
is given by
\begin{align*}
L^K_{(g,h)}(a,b)=(ga,hb).
\end{align*}
Therefore its differential at $e_K$ acts componentwise:
\begin{align*}
(dL^K_{(g,h)})_{e_K}(X,Y)=((dL^G_g)_{e_G}(X),(dL^H_h)_{e_H}(Y)).
\end{align*}
Hence
\begin{align*}
\widetilde{(X,Y)}_{(g,h)}=(\widetilde X_g,\widetilde Y_h).
\end{align*}
[guided]
The purpose of this step is to make the product identification compatible with the definition of the Lie bracket. The [Lie algebra](/page/Lie%20Algebra) bracket of a Lie group is defined by taking the bracket of the corresponding left-invariant vector fields and evaluating at the identity. So we must understand exactly which left-invariant vector field on $G\times H$ corresponds to a pair $(X,Y)$.
Let $X\in\mathfrak g=T_{e_G}G$ and $Y\in\mathfrak h=T_{e_H}H$. The left-invariant vector field on $G$ determined by $X$ is the smooth vector field $\widetilde X\in\mathfrak X(G)$ satisfying
\begin{align*}
\widetilde X_g=(dL^G_g)_{e_G}(X)
\end{align*}
for every $g\in G$, where $L^G_g:G\to G$ is the smooth map $a\mapsto ga$. Likewise, the left-invariant vector field on $H$ determined by $Y$ is the smooth vector field $\widetilde Y\in\mathfrak X(H)$ satisfying
\begin{align*}
\widetilde Y_h=(dL^H_h)_{e_H}(Y)
\end{align*}
for every $h\in H$, where $L^H_h:H\to H$ is the smooth map $b\mapsto hb$.
Now consider the product Lie group $K=G\times H$. Its left translation by $(g,h)\in K$ is the smooth map $L^K_{(g,h)}:K\to K$ defined by
\begin{align*}
L^K_{(g,h)}(a,b)=(ga,hb).
\end{align*}
This formula shows that $L^K_{(g,h)}$ is exactly the product map $L^G_g\times L^H_h$. Therefore its differential at the identity $e_K=(e_G,e_H)$ sends a pair of tangent vectors to the pair of differentials:
\begin{align*}
(dL^K_{(g,h)})_{e_K}(X,Y)=((dL^G_g)_{e_G}(X),(dL^H_h)_{e_H}(Y)).
\end{align*}
By the definitions of $\widetilde X$ and $\widetilde Y$, the right-hand side is
\begin{align*}
(\widetilde X_g,\widetilde Y_h).
\end{align*}
Thus the left-invariant vector field on $K$ determined by $(X,Y)$ is the product vector field
\begin{align*}
\widetilde{(X,Y)}_{(g,h)}=(\widetilde X_g,\widetilde Y_h).
\end{align*}
This is the key compatibility statement: the product tangent vector produces the product of the corresponding left-invariant vector fields.
[/guided]
[/step]
[step:Compute the bracket of product left-invariant vector fields componentwise]
Let $X_1,X_2\in\mathfrak g$ and $Y_1,Y_2\in\mathfrak h$. Let $\widetilde X_1,\widetilde X_2\in\mathfrak X(G)$ and $\widetilde Y_1,\widetilde Y_2\in\mathfrak X(H)$ be their corresponding left-invariant vector fields.
Define vector fields $A,B\in\mathfrak X(K)$ by
\begin{align*}
A_{(g,h)}=(\widetilde X_{1,g},\widetilde Y_{1,h})
\end{align*}
and
\begin{align*}
B_{(g,h)}=(\widetilde X_{2,g},\widetilde Y_{2,h})
\end{align*}
for every $(g,h)\in K$. By the previous step, $A=\widetilde{(X_1,Y_1)}$ and $B=\widetilde{(X_2,Y_2)}$.
Choose a product coordinate chart around $(g,h)$ with coordinates $(x_1,\dots,x_m,y_1,\dots,y_n)$, where $(x_1,\dots,x_m)$ are coordinates from a chart on $G$ and $(y_1,\dots,y_n)$ are coordinates from a chart on $H$. In these coordinates, there are smooth coefficient functions $a_i,b_i$ on the $G$-chart and $\alpha_j,\beta_j$ on the $H$-chart such that
\begin{align*}
A = \sum_{i=1}^{m} a_i(x)\,\partial_{x_i} + \sum_{j=1}^{n} \alpha_j(y)\,\partial_{y_j}
\end{align*}
and
\begin{align*}
B = \sum_{i=1}^{m} b_i(x)\,\partial_{x_i} + \sum_{j=1}^{n} \beta_j(y)\,\partial_{y_j}.
\end{align*}
Using the coordinate formula for the Lie bracket of vector fields, we obtain
\begin{align*}
[A,B]
=
\sum_{i=1}^{m}\bigl(A(b_i)-B(a_i)\bigr)\,\partial_{x_i}
+
\sum_{j=1}^{n}\bigl(A(\beta_j)-B(\alpha_j)\bigr)\,\partial_{y_j}.
\end{align*}
Because $a_i$ and $b_i$ depend only on the $G$-coordinates and $\alpha_j$ and $\beta_j$ depend only on the $H$-coordinates, the mixed derivatives vanish. Therefore the $G$-component of $[A,B]$ is $[\widetilde X_1,\widetilde X_2]$ and the $H$-component is $[\widetilde Y_1,\widetilde Y_2]$, so
\begin{align*}
[A,B]_{(g,h)}=([\widetilde X_1,\widetilde X_2]_g,[\widetilde Y_1,\widetilde Y_2]_h).
\end{align*}
Evaluating at $(e_G,e_H)$ gives
\begin{align*}
[A,B]_{e_K}=([\widetilde X_1,\widetilde X_2]_{e_G},[\widetilde Y_1,\widetilde Y_2]_{e_H}).
\end{align*}
[/step]
[step:Conclude that the product identification preserves the Lie bracket]
By definition of the Lie algebra bracket on a Lie group in terms of left-invariant vector fields,
\begin{align*}
[(X_1,Y_1),(X_2,Y_2)]_{\operatorname{Lie}(K)}=[\widetilde{(X_1,Y_1)},\widetilde{(X_2,Y_2)}]_{e_K}.
\end{align*}
Using the componentwise bracket computation from the previous step, this becomes
\begin{align*}
[(X_1,Y_1),(X_2,Y_2)]_{\operatorname{Lie}(K)}=([\widetilde X_1,\widetilde X_2]_{e_G},[\widetilde Y_1,\widetilde Y_2]_{e_H}).
\end{align*}
Again by the definition of the Lie algebra brackets on $G$ and $H$,
\begin{align*}
[\widetilde X_1,\widetilde X_2]_{e_G}=[X_1,X_2]_{\mathfrak g}
\end{align*}
and
\begin{align*}
[\widetilde Y_1,\widetilde Y_2]_{e_H}=[Y_1,Y_2]_{\mathfrak h}.
\end{align*}
Therefore
\begin{align*}
[(X_1,Y_1),(X_2,Y_2)]_{\operatorname{Lie}(G\times H)}=([X_1,X_2]_{\mathfrak g},[Y_1,Y_2]_{\mathfrak h}).
\end{align*}
Thus the natural vector-space isomorphism
\begin{align*}
\operatorname{Lie}(G\times H)\cong\mathfrak g\oplus\mathfrak h
\end{align*}
is an isomorphism of Lie algebras, where $\mathfrak g\oplus\mathfrak h$ has the componentwise bracket.
[/step]
