[proofplan] We split the [Lie algebra](/page/Lie%20Algebra) $\mathfrak g$ of $G$ into its central abelian part and its semisimple derived part using an $\operatorname{Ad}(G)$-invariant [inner product](/page/Inner%20Product). The central summand integrates inside $G$ to the identity component of the center, which is a torus, while the derived summand is integrated by a simply connected compact semisimple Lie group $K$. Multiplication gives a homomorphism from the product of these two factors onto $G$ whose differential is an isomorphism, so its kernel is discrete. Since the source is compact and connected, the kernel is finite and central, and centrality identifies it as a subgroup of $T^k\times Z(K)$. [/proofplan]

[step:Split the compact Lie algebra into its central and semisimple summands]Let $\mathfrak g:=\operatorname{Lie}(G)$ denote the Lie algebra of $G$, and let \begin{align*} \mathfrak z:=\mathfrak z(\mathfrak g):=\{X\in\mathfrak g:[X,Y]=0\text{ for every }Y\in\mathfrak g\} \end{align*} be its center. Let \begin{align*} \mathfrak s:=[\mathfrak g,\mathfrak g] \end{align*} denote the derived Lie subalgebra, the real linear span of all brackets $[X,Y]$ with $X,Y\in\mathfrak g$. Because $G$ is compact, averaging any real inner product on $\mathfrak g$ over $G$ with respect to normalized Haar probability measure gives an $\operatorname{Ad}(G)$-invariant real inner product $(\cdot,\cdot)_{\mathfrak g}$ on $\mathfrak g$ (citing a result not yet in the wiki: Existence of Ad-invariant inner product on compact Lie algebra). Its infinitesimal invariance says that, for all $X,Y,Z\in\mathfrak g$, \begin{align*} ([X,Y],Z)_{\mathfrak g}+(Y,[X,Z])_{\mathfrak g}=0. \end{align*} If $Z\in\mathfrak z$, then $[X,Z]=0$ for every $X\in\mathfrak g$, and therefore \begin{align*} ([X,Y],Z)_{\mathfrak g}=0 \end{align*} for every $X,Y\in\mathfrak g$. Hence $\mathfrak s\subseteq \mathfrak z^\perp$. Conversely, if $U\in\mathfrak s^\perp$, then for every $X,Y\in\mathfrak g$, \begin{align*} ([U,X],Y)_{\mathfrak g}=-(X,[U,Y])_{\mathfrak g}. \end{align*} Since $[U,Y]\in\mathfrak s$ and $X$ is arbitrary, this identity alone does not yet force $[U,X]=0$. We use the standard compact Lie algebra structure theorem instead: a compact Lie algebra decomposes orthogonally as the [direct sum](/page/Direct%20Sum) of its center and its derived semisimple ideal (citing a result not yet in the wiki: Compact Lie algebra decomposition into center plus semisimple derived algebra). Applied to $\mathfrak g$, it gives \begin{align*} \mathfrak g=\mathfrak z\oplus\mathfrak s \end{align*} as an orthogonal direct sum of ideals, and $\mathfrak s$ is a compact semisimple Lie algebra.[/step]

[guided]We first isolate the two Lie-algebraic pieces that should become the two group factors. Let \begin{align*} \mathfrak g:=\operatorname{Lie}(G) \end{align*} denote the Lie algebra of $G$, let \begin{align*} \mathfrak z:=\mathfrak z(\mathfrak g):=\{X\in\mathfrak g:[X,Y]=0\text{ for every }Y\in\mathfrak g\} \end{align*} be its center, and let \begin{align*} \mathfrak s:=[\mathfrak g,\mathfrak g] \end{align*} denote the derived Lie subalgebra. The point is that a compact connected Lie group splits into a central abelian part and a semisimple commutator part, and these are the two Lie-algebraic pieces that will integrate to the factors $T^k$ and $K$. To make that splitting precise, we use compactness. Averaging any real inner product on $\mathfrak g$ over $G$ with respect to normalized Haar probability measure gives an $\operatorname{Ad}(G)$-invariant real inner product $(\cdot,\cdot)_{\mathfrak g}$ on $\mathfrak g$ (citing a result not yet in the wiki: Existence of Ad-invariant inner product on compact Lie algebra). Infinitesimal invariance says that for all $X,Y,Z\in\mathfrak g$, \begin{align*} ([X,Y],Z)_{\mathfrak g}+(Y,[X,Z])_{\mathfrak g}=0. \end{align*} Now take $Z\in\mathfrak z$. Then $[X,Z]=0$ for every $X\in\mathfrak g$, so for every $X,Y\in\mathfrak g$, \begin{align*} ([X,Y],Z)_{\mathfrak g}=0. \end{align*} This means that every bracket is orthogonal to the center, and therefore \begin{align*} \mathfrak s\subseteq\mathfrak z^\perp. \end{align*} The reverse inclusion is not forced by the displayed orthogonality identity alone. Indeed, if $U\in\mathfrak s^\perp$, then for every $X,Y\in\mathfrak g$ we have \begin{align*} ([U,X],Y)_{\mathfrak g}=-(X,[U,Y])_{\mathfrak g}. \end{align*} Since $[U,Y]\in\mathfrak s$ and $X$ is arbitrary, this relation by itself does not yet imply that $[U,X]=0$. At this point we invoke the standard compact Lie algebra decomposition theorem: a compact Lie algebra decomposes orthogonally as the direct sum of its center and its derived semisimple ideal (citing a result not yet in the wiki: Compact Lie algebra decomposition into center plus semisimple derived algebra). Applied to $\mathfrak g$, it gives \begin{align*} \mathfrak g=\mathfrak z\oplus\mathfrak s \end{align*} as an orthogonal direct sum of ideals, and $\mathfrak s=[\mathfrak g,\mathfrak g]$ is a compact semisimple Lie algebra. This is the Lie-algebraic splitting that the rest of the proof integrates to the global product decomposition.[/guided]

[step:Integrate the central summand as a torus inside $G$] Let $Z(G)$ denote the center of $G$, and define \begin{align*} C:=Z(G)^\circ \end{align*} to be the identity component of $Z(G)$. Since $Z(G)$ is a closed subgroup of the compact Lie group $G$, the subgroup $C$ is a compact connected abelian Lie group. Therefore there exists an integer $k\geq 0$ and an isomorphism of Lie groups \begin{align*} \alpha:T^k\to C. \end{align*} Moreover, \begin{align*} \operatorname{Lie}(C)=\mathfrak z. \end{align*} Indeed, the Lie algebra of $Z(G)$ consists exactly of the elements $X\in\mathfrak g$ whose one-parameter subgroup $\exp_G(tX)$ commutes with every element of $G$; because $G$ is connected, this is equivalent to $[X,Y]=0$ for every $Y\in\mathfrak g$. Thus the central summand $\mathfrak z$ is integrated by the compact torus $C\cong T^k$ inside $G$. [/step]

[step:Integrate the semisimple summand by a simply connected compact group] Since $\mathfrak s$ is a compact semisimple Lie algebra, the standard integration theorem for compact semisimple Lie algebras gives a simply connected compact semisimple Lie group $K$ with \begin{align*} \operatorname{Lie}(K)\cong\mathfrak s \end{align*} (citing a result not yet in the wiki: Existence and compactness of the simply connected Lie group integrating a compact semisimple Lie algebra). Fix a Lie algebra isomorphism \begin{align*} \theta:\operatorname{Lie}(K)\to\mathfrak s. \end{align*} Composing $\theta$ with the inclusion $\mathfrak s\hookrightarrow\mathfrak g$ gives a Lie algebra homomorphism from $\operatorname{Lie}(K)$ to $\mathfrak g$. Since $K$ is simply connected, [Lie's second theorem](/theorems/8809) integrates this Lie algebra homomorphism to a unique Lie [group homomorphism](/page/Group%20Homomorphism) \begin{align*} \beta:K\to G \end{align*} whose differential at the identity is the inclusion $\mathfrak s\hookrightarrow\mathfrak g$ under the identification $\theta$ (citing a result not yet in the wiki: Lie algebra homomorphisms from simply connected Lie groups integrate uniquely). [/step]

[step:Construct the product homomorphism with isomorphic differential] Define a map \begin{align*} \Phi:T^k\times K\to G \end{align*} by \begin{align*} \Phi(a,u):=\alpha(a)\beta(u) \end{align*} for $a\in T^k$ and $u\in K$. Since $\alpha(T^k)=C\subseteq Z(G)$, every element $\alpha(a)$ commutes with every element of $\beta(K)$. Therefore $\Phi$ is a Lie group homomorphism: \begin{align*} \Phi(a,u)\Phi(b,v)=\alpha(a)\beta(u)\alpha(b)\beta(v)=\alpha(ab)\beta(uv)=\Phi(ab,uv). \end{align*} Let $\mathfrak t_k:=\operatorname{Lie}(T^k)$. The differential of $\alpha$ at the identity is an isomorphism \begin{align*} d\alpha_e:\mathfrak t_k\to\mathfrak z. \end{align*} Using the product identification \begin{align*} \operatorname{Lie}(T^k\times K)=\mathfrak t_k\oplus\operatorname{Lie}(K), \end{align*} the differential of $\Phi$ at the identity is \begin{align*} d\Phi_{(e,e)}(A,B)=d\alpha_e(A)+\theta(B). \end{align*} Because \begin{align*} \mathfrak g=\mathfrak z\oplus\mathfrak s, \end{align*} this differential is a real-linear isomorphism from $\mathfrak t_k\oplus\operatorname{Lie}(K)$ onto $\mathfrak g$. [/step]

[step:Show the product homomorphism is surjective]Since $d\Phi_{(e,e)}$ is an isomorphism, the [inverse function theorem](/theorems/51) for smooth manifolds implies that $\Phi$ maps a neighbourhood of $(e,e)\in T^k\times K$ onto a neighbourhood of $e\in G$. Because $\Phi$ is a homomorphism, its image \begin{align*} H:=\Phi(T^k\times K) \end{align*} is a subgroup of $G$ containing a neighbourhood of $e$. Translating this neighbourhood by elements of $H$ shows that $H$ is open in $G$. The subgroup $H$ is also closed because $T^k\times K$ is compact and $\Phi$ is continuous, so $H$ is compact in the [Hausdorff space](/page/Hausdorff%20Space) $G$. Since $G$ is connected and $H$ is a nonempty subset that is both open and closed, one has \begin{align*} H=G. \end{align*} Thus $\Phi$ is surjective.[/step]

[guided]The differential computation tells us that $\Phi$ is locally an isomorphism near the identity, but we need a global statement: every element of $G$ must be a product of a central torus element and a semisimple element. Since $d\Phi_{(e,e)}$ is an isomorphism, the inverse function theorem applies to the smooth map $\Phi:T^k\times K\to G$. It gives open neighbourhoods $U\subseteq T^k\times K$ of $(e,e)$ and $V\subseteq G$ of $e$ such that \begin{align*} \Phi(U)=V. \end{align*} In particular, the image \begin{align*} H:=\Phi(T^k\times K) \end{align*} contains an open neighbourhood of the identity element of $G$. Because $\Phi$ is a group homomorphism, $H$ is a subgroup of $G$. If $h\in H$, then left multiplication by $h$ maps the neighbourhood $V$ homeomorphically onto the open neighbourhood $hV$ of $h$, and $hV\subseteq H$ because $H$ is a subgroup. Therefore every point of $H$ has an open neighbourhood contained in $H$, so $H$ is open in $G$. We also need a closedness argument. The product $T^k\times K$ is compact because both factors are compact. Since $\Phi$ is continuous, the image $H$ is compact. The Lie group $G$ is Hausdorff, so compact subsets of $G$ are closed. Hence $H$ is closed in $G$. Now use connectedness. The subset $H$ is nonempty because it contains $e$, and it is both open and closed in the [connected space](/page/Connected%20Space) $G$. Therefore \begin{align*} H=G. \end{align*} This proves that $\Phi$ is surjective.[/guided]

[step:Identify the kernel as a finite central subgroup] Define \begin{align*} F:=\ker\Phi=\{(a,u)\in T^k\times K:\alpha(a)\beta(u)=e\}. \end{align*} Because $d\Phi_{(e,e)}$ is an isomorphism, the Lie algebra of $F$ is \begin{align*} \ker d\Phi_{(e,e)}=\{0\}. \end{align*} Hence $F$ is a discrete Lie subgroup of $T^k\times K$. Since $T^k\times K$ is compact, every discrete closed subgroup is finite (citing a result not yet in the wiki: Discrete subgroup of a compact group is finite). Thus $F$ is finite. The kernel of a homomorphism is normal, so $F\trianglelefteq T^k\times K$. We now show it is central. For a fixed element $f\in F$, define the conjugation-orbit map \begin{align*} \gamma_f:T^k\times K\to F \end{align*} by \begin{align*} \gamma_f(x):=xfx^{-1}. \end{align*} The map $\gamma_f$ is continuous because multiplication and inversion are continuous. Its domain $T^k\times K$ is connected, while its codomain $F$ is discrete. Therefore $\gamma_f$ is constant. Since $\gamma_f(e,e)=f$, it follows that \begin{align*} xfx^{-1}=f \end{align*} for every $x\in T^k\times K$. Hence $F\leq Z(T^k\times K)$. Finally, \begin{align*} Z(T^k\times K)=T^k\times Z(K), \end{align*} because $T^k$ is abelian and the center of a direct product is the product of the centers. Therefore \begin{align*} F\leq T^k\times Z(K). \end{align*} [/step]

[step:Pass to the quotient and obtain the required isomorphism] Since $\Phi:T^k\times K\to G$ is a surjective Lie group homomorphism with kernel $F$, the [first isomorphism theorem](/theorems/791) for Lie groups gives an isomorphism of Lie groups \begin{align*} (T^k\times K)/F\cong G. \end{align*} The subgroup $F$ is finite and central in $T^k\times K$, and the previous step shows that \begin{align*} F\leq T^k\times Z(K). \end{align*} Thus $G$ has the asserted form \begin{align*} G\cong (T^k\times K)/F, \end{align*} where $K$ is simply connected, compact, and semisimple. This proves the theorem. [/step]