[proofplan] We first show that every element of $G$ lies in the image of the Lie exponential map $\exp:\mathfrak g\to G$ by using a bi-invariant Riemannian metric and the fact that geodesics through the identity are one-parameter subgroups. Given such an exponential representative $g=\exp X$, we enlarge the line spanned by $X$ to a maximal abelian Lie subalgebra of $\mathfrak g$. The closure of the exponential image of that subalgebra is then a maximal torus containing the original element. Finally, the [conjugacy theorem for maximal tori](/theorems/9720) moves this maximal torus onto the fixed maximal torus $T$, carrying the element into $T$. [/proofplan]

[step:Realize the given element as an exponential]Let $\mathfrak g := T_eG$ denote the [Lie algebra](/page/Lie%20Algebra) of $G$, where $e \in G$ is the identity element. Define the adjoint representation by \begin{align*} \operatorname{Ad}: G \to GL(\mathfrak g),\quad h \mapsto \operatorname{Ad}(h). \end{align*} Write \begin{align*} \operatorname{Ad}(G) := \{\operatorname{Ad}(h) : h \in G\} \subseteq GL(\mathfrak g). \end{align*} Choose an [inner product](/page/Inner%20Product) $(\cdot,\cdot)_0$ on the real [vector space](/page/Vector%20Space) $\mathfrak g$. Let $\mu_G$ denote the normalized Haar probability measure on $G$. Since $G$ is compact, averaging $(\cdot,\cdot)_0$ against $\mu_G$ over the adjoint action of $G$ gives an $\operatorname{Ad}(G)$-invariant inner product $(\cdot,\cdot)_{\mathfrak g}$ on $\mathfrak g$; positivity is preserved because the average of positive numbers is positive, and invariance follows from the left-invariance of $\mu_G$. Equivalently, this determines a bi-invariant Riemannian metric on $G$. We use the following standard Riemannian fact, not yet cited from the wiki: since $G$ is compact and connected, compactness gives metric completeness and connectedness places $e$ and $g$ in the same [connected component](/page/Connected%20Component), so the Hopf-Rinow theorem gives a minimizing geodesic from $e$ to $g$. Thus, for the chosen element $g\in G$, there exists a geodesic \begin{align*} \gamma:[0,1]\to G \end{align*} with $\gamma(0)=e$ and $\gamma(1)=g$. Define \begin{align*} X := \gamma'(0). \end{align*} Then $X\in T_eG=\mathfrak g$. Let $\exp:\mathfrak g\to G$ denote the Lie exponential map. For a bi-invariant metric on a Lie group, the geodesic with initial value $e$ and initial velocity $X$ is the one-parameter subgroup \begin{align*} c_X:\mathbb R \to G \end{align*} given by $c_X(s)=\exp(sX)$ for every $s\in\mathbb R$. Indeed, if $\widetilde X$ is the left-invariant vector field on $G$ determined by $X$, then bi-invariance gives \begin{align*} \nabla_{\widetilde X}\widetilde X=\frac{1}{2}[\widetilde X,\widetilde X]=0. \end{align*} Thus the integral curve of $\widetilde X$ through $e$ is a geodesic. By uniqueness of geodesics with prescribed initial point and velocity, $\gamma(s)=\exp(sX)$ for all $s\in[0,1]$. Evaluating at $s=1$ gives \begin{align*} g=\exp X. \end{align*}[/step]

[guided]The first task is to place $g$ inside some torus. A practical way to do this is to prove that $g$ is an exponential. Once $g=\exp X$, any abelian Lie subalgebra containing $X$ will exponentiate to an abelian subgroup containing $g$. Let $\mathfrak g:=T_eG$ be the Lie algebra of $G$. Choose an arbitrary inner product $(\cdot,\cdot)_0$ on $\mathfrak g$. Let $\mu_G$ denote the normalized Haar probability measure on $G$. Because $G$ is compact, we can average this inner product over the adjoint action of $G$ with respect to $\mu_G$ to obtain an $\operatorname{Ad}(G)$-invariant inner product $(\cdot,\cdot)_{\mathfrak g}$. Positivity is preserved because the average of positive numbers is positive, and $\operatorname{Ad}(G)$-invariance is exactly the condition needed to extend the form by left translation to a bi-invariant Riemannian metric on $G$. We now use the Hopf-Rinow theorem, a standard Riemannian result not yet cited from the wiki. Since $G$ is compact, it is complete for the chosen Riemannian metric, and since $G$ is connected, the points $e$ and $g$ lie in the same connected component. Therefore Hopf-Rinow gives a geodesic \begin{align*} \gamma:[0,1]\to G \end{align*} such that $\gamma(0)=e$ and $\gamma(1)=g$. Define \begin{align*} X := \gamma'(0). \end{align*} Then $X\in T_eG=\mathfrak g$. Why does this imply that $g$ is an exponential? The point of choosing a bi-invariant metric is that its geodesics through the identity are exactly one-parameter subgroups. More explicitly, let $\widetilde X$ be the left-invariant vector field on $G$ whose value at $e$ is $X$. For a bi-invariant metric, the Levi-Civita connection satisfies \begin{align*} \nabla_{\widetilde X}\widetilde X=\frac{1}{2}[\widetilde X,\widetilde X]. \end{align*} The Lie bracket of a vector field with itself is zero, so \begin{align*} \nabla_{\widetilde X}\widetilde X=0. \end{align*} Therefore the integral curve of $\widetilde X$ through $e$ is a geodesic. Let $\exp:\mathfrak g\to G$ denote the Lie exponential map. That integral curve is the one-parameter subgroup \begin{align*} c_X:\mathbb R \to G \end{align*} given by $c_X(s)=\exp(sX)$ for every $s\in\mathbb R$. Both $\gamma$ and $c_X|_{[0,1]}$ are geodesics with the same initial point $e$ and the same initial velocity $X$. By uniqueness for the geodesic equation, they agree on $[0,1]$. Hence \begin{align*} g=\gamma(1)=c_X(1)=\exp X. \end{align*}[/guided]

[step:Place the exponential inside a maximal torus]Let $\mathbb R X\le \mathfrak g$ denote the real linear span of $X$. Since $[\mathbb R X,\mathbb R X]=0$, this is an abelian Lie subalgebra of $\mathfrak g$. Because $\mathfrak g$ is finite-dimensional, there exists a maximal abelian Lie subalgebra $\mathfrak a\le \mathfrak g$ such that \begin{align*} \mathbb R X\subset \mathfrak a. \end{align*} Define \begin{align*} S:=\overline{\exp(\mathfrak a)}\le G. \end{align*} Because $G$ is compact and connected by hypothesis, and $\mathfrak a$ is a maximal abelian Lie subalgebra of $\mathfrak g$, [citetheorem:9718] applies and shows that $S$ is a maximal torus of $G$ and that its Lie algebra is $\mathfrak a$. Since $X\in\mathfrak a$, we have \begin{align*} g = \exp X \in \exp(\mathfrak a) \subset S. \end{align*} Thus $g$ lies in the maximal torus $S$.[/step]

[guided]The aim of this step is to place $g$ inside a torus that is large enough to be maximal. Starting from the logarithm $X$ with $g=\exp X$, we consider the one-dimensional abelian Lie subalgebra $\mathbb R X\le \mathfrak g$. Since $\mathfrak g$ is finite-dimensional, we can extend $\mathbb R X$ to a maximal abelian Lie subalgebra $\mathfrak a\le \mathfrak g$. Now set \begin{align*} S:=\overline{\exp(\mathfrak a)}\le G. \end{align*} The closure is taken because $\exp(\mathfrak a)$ need not itself be closed, but its closure is a torus. The reason for invoking [citetheorem:9718] is that it identifies the exponentiated image of a maximal abelian Lie subalgebra inside a compact connected Lie group as a maximal torus. The theorem applies here because $G$ is compact and connected by hypothesis, and $\mathfrak a$ is maximal abelian by construction. Therefore $S$ is a maximal torus of $G$ with Lie algebra $\mathfrak a$. Since $X\in\mathfrak a$, the exponential $\exp X$ belongs to $\exp(\mathfrak a)$ and hence to $S$. Because $g=\exp X$, this proves that the given element lies in a maximal torus.[/guided]

[step:Conjugate that maximal torus onto the fixed maximal torus]The subgroup $S\le G$ constructed above is a maximal torus of $G$, and $T\le G$ is a maximal torus by hypothesis. By [citetheorem:9720], there exists an element $h\in G$ such that \begin{align*} hSh^{-1}=T. \end{align*} Since $g\in S$, it follows that \begin{align*} hgh^{-1} \in hSh^{-1} = T. \end{align*} Setting $t:=hgh^{-1}$ gives $t\in T$ and \begin{align*} hgh^{-1}=t. \end{align*} Because $g\in G$ was arbitrary, every element of $G$ is conjugate to an element of the fixed maximal torus $T$.[/step]

[guided]This final step uses only the conjugacy theorem for maximal tori. We have already shown that the arbitrary element $g$ lies in some maximal torus $S\le G$, and the theorem says that any two maximal tori in a compact connected Lie group are conjugate. Because $G$ is compact and connected by hypothesis, [citetheorem:9720] applies to $S$ and the fixed maximal torus $T$, producing an element $h\in G$ with \begin{align*} hSh^{-1}=T. \end{align*} Conjugating the inclusion $g\in S$ by $h$ gives $hgh^{-1}\in T$. Writing $t:=hgh^{-1}$ yields the required element of $T$ conjugate to $g$.[/guided]