[proofplan] We first show that $\exp(\mathfrak t)$ is a connected abelian subgroup of $G$, so its closure $T$ is a compact connected abelian subgroup. The closed subgroup theorem makes $T$ a Lie subgroup, and the structure theorem for compact connected abelian Lie groups identifies it as a torus. The one-parameter [subgroup criterion](/theorems/932) gives $\mathfrak t\subseteq \operatorname{Lie}(T)$, while abelianness of $T$ forces $\operatorname{Lie}(T)$ to be abelian, so maximality gives equality. Finally, any larger torus would have the same [Lie algebra](/page/Lie%20Algebra) and therefore coincide with $T$. [/proofplan] [step:Show that the exponential image is a connected abelian subgroup] Let \begin{align*} H:=\exp(\mathfrak t)=\{\exp X:X\in\mathfrak t\}\subset G. \end{align*} Since $\mathfrak t$ is an abelian Lie algebra, for all $X,Y\in\mathfrak t$ the Baker-Campbell-Hausdorff formula reduces to \begin{align*} \exp X\exp Y=\exp(X+Y). \end{align*} Also $(\exp X)^{-1}=\exp(-X)$. Hence $H$ is a subgroup of $G$. The restricted exponential map $\exp|_{\mathfrak t}:\mathfrak t\to G$ is continuous, and $\mathfrak t$ is connected as a finite-dimensional real [vector space](/page/Vector%20Space). Therefore $H=\exp(\mathfrak t)$ is connected. The same displayed product formula also shows that $H$ is abelian. [guided] We need a subgroup whose closure will become the candidate torus. Define \begin{align*} H:=\exp(\mathfrak t)=\{\exp X:X\in\mathfrak t\}\subset G. \end{align*} The point of assuming $\mathfrak t$ is abelian is that the exponential map behaves additively on $\mathfrak t$. For $X,Y\in\mathfrak t$, the Lie bracket satisfies $[X,Y]=0$, so the Baker-Campbell-Hausdorff formula has no commutator terms and gives \begin{align*} \exp X\exp Y=\exp(X+Y). \end{align*} Because $X+Y\in\mathfrak t$, the product of two elements of $H$ lies again in $H$. Also $-X\in\mathfrak t$, and \begin{align*} (\exp X)^{-1}=\exp(-X), \end{align*} so inverses also lie in $H$. Thus $H$ is a subgroup of $G$. The subgroup is connected because it is the continuous image of a [connected space](/page/Connected%20Space). More precisely, the restricted exponential map $\exp|_{\mathfrak t}:\mathfrak t\to G$ is continuous, and $\mathfrak t$ is connected as a real vector space. Therefore its image $H$ is connected. Finally, the same identity \begin{align*} \exp X\exp Y=\exp(X+Y)=\exp(Y+X)=\exp Y\exp X \end{align*} shows that any two elements of $H$ commute. Hence $H$ is a connected abelian subgroup of $G$. [/guided] [/step] [step:Pass to the closure and identify it as a torus] By definition, \begin{align*} T=\overline H. \end{align*} We first verify that $T$ is a subgroup of $G$. Let \begin{align*} m:G\times G&\to G \end{align*} be the multiplication map, and let \begin{align*} \iota:G&\to G \end{align*} be the inversion map. Both maps are continuous because $G$ is a Lie group. Since $H$ is a subgroup, $m(H\times H)\subset H$ and $\iota(H)=H$. Therefore \begin{align*} m(T\times T)=m(\overline H\times \overline H)=m(\overline{H\times H})\subseteq \overline{m(H\times H)}\subseteq \overline H=T, \end{align*} and \begin{align*} \iota(T)=\iota(\overline H)\subseteq \overline{\iota(H)}=\overline H=T. \end{align*} Thus $T$ is a subgroup of $G$. Since $G$ is compact, the closed subset $T\subset G$ is compact. Since $H$ is connected, its closure $T$ is connected. Since $H$ is abelian and the commutator map \begin{align*}c:G\times G\to G,\quad (a,b)\mapsto aba^{-1}b^{-1}\end{align*} is continuous, $c$ is equal to the identity element $e\in G$ on $H\times H$ and therefore on $\overline{H\times H}=T\times T$. Thus $T$ is abelian. By the closed subgroup theorem, whose subgroup hypothesis has just been verified and whose closedness hypothesis holds by the definition of $T$, $T$ is an embedded Lie subgroup of $G$. By the structure theorem for compact connected abelian Lie groups, every compact connected abelian Lie group is isomorphic to a finite-dimensional torus. Therefore $T$ is a torus. [guided] The definition $T=\overline H$ gives a closed subset of $G$, but the closed subgroup theorem cannot be applied until we know that this closed subset is actually a subgroup. We prove this directly from continuity of the group operations. Let \begin{align*} m:G\times G&\to G \end{align*} be the multiplication map, and let \begin{align*} \iota:G&\to G \end{align*} be the inversion map. These maps are continuous because $G$ is a Lie group. Since $H$ is already a subgroup, multiplying two elements of $H$ stays in $H$, and inverting an element of $H$ stays in $H$. Now use the general closure rule for continuous maps: the image of the closure of a set is contained in the closure of the image. Also, in the [product topology](/page/Product%20Topology) one has $\overline H\times\overline H=\overline{H\times H}$. Hence \begin{align*} m(T\times T)=m(\overline H\times \overline H)=m(\overline{H\times H})\subseteq \overline{m(H\times H)}\subseteq \overline H=T. \end{align*} This proves closure of $T$ under multiplication. Similarly, \begin{align*} \iota(T)=\iota(\overline H)\subseteq \overline{\iota(H)}=\overline H=T, \end{align*} so $T$ is closed under inverses. Since the identity element belongs to $H\subset T$, we have proved that $T$ is a subgroup of $G$. With the subgroup point settled, the remaining topological properties follow from standard closure arguments. The set $T$ is compact because it is closed in the compact Lie group $G$. It is connected because it is the closure of the connected subset $H$. To prove abelianness, define the commutator map \begin{align*} c:G\times G&\to G \end{align*} by \begin{align*} c(a,b)=aba^{-1}b^{-1}. \end{align*} The map $c$ is continuous, and because $H$ is abelian it takes the constant value $e$ on $H\times H$. Since $T\times T=\overline{H\times H}$, continuity forces $c(a,b)=e$ for every $(a,b)\in T\times T$. Thus every two elements of $T$ commute. We may now apply the closed subgroup theorem: $T$ is a closed subgroup of the Lie group $G$, so $T$ is an embedded Lie subgroup of $G$. The hypotheses of the structure theorem for compact connected abelian Lie groups are also verified: $T$ is compact, connected, abelian, and a Lie group. Therefore $T$ is isomorphic to a finite-dimensional torus. [/guided] [/step] [step:Compute the Lie algebra of the closure] Let $\mathfrak s:=\operatorname{Lie}(T)\leq\mathfrak g$. For each $X\in\mathfrak t$, define the one-parameter subgroup \begin{align*} \gamma_X:\mathbb R&\to G \end{align*} \begin{align*} r&\mapsto \exp(rX). \end{align*} Since $rX\in\mathfrak t$ for every $r\in\mathbb R$, we have $\gamma_X(\mathbb R)\subset H\subset T$. By the one-parameter subgroup criterion for the Lie algebra of a Lie subgroup, $X=\gamma_X'(0)$ lies in $\mathfrak s$. Hence \begin{align*} \mathfrak t\subseteq \mathfrak s. \end{align*} Because $T$ is abelian, its Lie algebra $\mathfrak s$ is abelian. Since $\mathfrak t$ is maximal among abelian Lie subalgebras of $\mathfrak g$, the inclusion $\mathfrak t\subseteq\mathfrak s$ forces \begin{align*} \operatorname{Lie}(T)=\mathfrak s=\mathfrak t. \end{align*} [/step] [step:Exclude containment in a larger torus] Suppose $T'\leq G$ is a torus with $T\subseteq T'$. Let \begin{align*} \mathfrak t':=\operatorname{Lie}(T')\leq\mathfrak g. \end{align*} Since $T'$ is abelian, $\mathfrak t'$ is an abelian Lie subalgebra of $\mathfrak g$. The inclusion $T\subseteq T'$ gives \begin{align*} \operatorname{Lie}(T)\subseteq \operatorname{Lie}(T'), \end{align*} so \begin{align*} \mathfrak t\subseteq \mathfrak t'. \end{align*} By maximality of $\mathfrak t$, we get $\mathfrak t'=\mathfrak t$. It remains to justify that this equality of Lie algebras gives equality of tori. Since $T$ and $T'$ are connected Lie subgroups and $T\subseteq T'$, the inclusion \begin{align*} i:T\to T' \end{align*} has differential at the identity \begin{align*} di_e:\operatorname{Lie}(T)\to\operatorname{Lie}(T') \end{align*} equal to the identity map on $\mathfrak t$. Thus $di_e$ is an isomorphism. By the [inverse function theorem](/theorems/51) for Lie groups, $i(T)$ contains a neighbourhood of the identity in $T'$. Therefore $T$ is an open subgroup of $T'$. It is also closed in $T'$ because $T$ is compact and $T'$ is Hausdorff. Since $T'$ is connected, its only nonempty subset that is both open and closed and is a subgroup containing the identity component is the whole group. Hence \begin{align*} T=T'. \end{align*} Therefore $T$ is not properly contained in any larger torus, so $T$ is a maximal torus of $G$. [/step]