[proofplan] We let $T_1$ act by left multiplication on the homogeneous space $G/T_2$. After averaging an arbitrary Riemannian metric over the compact torus $T_1$, this action becomes isometric. The quotient $G/T_2$ is compact and connected, and the standard Euler characteristic theorem for maximal tori gives $\chi(G/T_2)\ne 0$, so the torus fixed point theorem gives a fixed coset. Translating the fixed coset condition gives an inclusion $g^{-1}T_1g\le T_2$, and maximality of the two tori upgrades the inclusion to equality. [/proofplan] [step:Make $T_1$ act smoothly on the compact connected manifold $G/T_2$] Let $M:=G/T_2$ be the left coset space, and let \begin{align*} q:G&\to M \end{align*} \begin{align*} g&\mapsto gT_2 \end{align*} be the quotient map. Since $T_2$ is a closed Lie subgroup of $G$, the quotient $M$ is a smooth compact manifold and $q$ is a smooth submersion. Compactness follows because $q$ is continuous and $G$ is compact. Connectedness follows because $q(G)=M$ and the continuous image of the [connected space](/page/Connected%20Space) $G$ is connected. Define the left action \begin{align*} \alpha:T_1\times M&\to M \end{align*} \begin{align*} (t,gT_2)&\mapsto tgT_2. \end{align*} This map is well-defined because if $gT_2=g'T_2$, then $g'=gh$ for some $h\in T_2$, hence $tg'T_2=tghT_2=tgT_2$. The action is smooth because it is induced by the smooth multiplication map $T_1\times G\to G$ and the quotient map $q$. [/step] [step:Average a Riemannian metric to make the action isometric] Choose a smooth Riemannian metric $h$ on $M$. Let $\mu_{T_1}$ denote the normalized Haar probability measure on the compact Lie group $T_1$. For $x\in M$ and $u,v\in T_xM$, define \begin{align*} \bar h_x(u,v):=\int_{T_1} h_{a^{-1}x}\bigl(d\alpha_{a^{-1},x}(u),d\alpha_{a^{-1},x}(v)\bigr)\,d\mu_{T_1}(a), \end{align*} where \begin{align*} \alpha_a:M&\to M \end{align*} \begin{align*} x&\mapsto \alpha(a,x) \end{align*} is the diffeomorphism determined by $a\in T_1$, and \begin{align*} d\alpha_{a^{-1},x}:T_xM&\to T_{a^{-1}x}M \end{align*} is its differential at $x$. The integrand is a smooth positive definite symmetric [bilinear form](/page/Bilinear%20Form) in $u$ and $v$, and averaging over the probability measure $\mu_{T_1}$ preserves symmetry and positive definiteness. Hence $\bar h$ is a smooth Riemannian metric on $M$. The left invariance of Haar measure gives, for every $b\in T_1$, \begin{align*} \bar h_{bx}\bigl(d\alpha_{b,x}(u),d\alpha_{b,x}(v)\bigr)=\bar h_x(u,v). \end{align*} Thus each diffeomorphism $\alpha_b:M\to M$ is an isometry of $(M,\bar h)$. [guided] We need an isometric torus action in order to use the fixed point theorem. The given action of $T_1$ on $M=G/T_2$ is smooth, but an arbitrary Riemannian metric need not be preserved by that action. The standard remedy is averaging over the compact acting group. Start with any smooth Riemannian metric $h$ on $M$. Let $\mu_{T_1}$ be normalized Haar probability measure on $T_1$. For each $x\in M$ and tangent vectors $u,v\in T_xM$, define a new bilinear form by \begin{align*} \bar h_x(u,v):=\int_{T_1} h_{a^{-1}x}\bigl(d\alpha_{a^{-1},x}(u),d\alpha_{a^{-1},x}(v)\bigr)\,d\mu_{T_1}(a). \end{align*} This formula pulls $u$ and $v$ back by the diffeomorphism $\alpha_{a^{-1}}$ before measuring them with $h$, then averages over all $a\in T_1$. For each fixed $a\in T_1$, the expression \begin{align*} (u,v)\mapsto h_{a^{-1}x}\bigl(d\alpha_{a^{-1},x}(u),d\alpha_{a^{-1},x}(v)\bigr) \end{align*} is a positive definite symmetric bilinear form on $T_xM$, because $d\alpha_{a^{-1},x}$ is a linear isomorphism and $h_{a^{-1}x}$ is positive definite. Integrating such forms against the probability measure $\mu_{T_1}$ again gives a positive definite symmetric bilinear form. Smoothness follows from smooth dependence of the action and differential on $(a,x)$ and compactness of $T_1$. Now verify invariance. Fix $b\in T_1$. For $x\in M$ and $u,v\in T_xM$, the definition gives \begin{align*} \bar h_{bx}\bigl(d\alpha_{b,x}(u),d\alpha_{b,x}(v)\bigr) =\int_{T_1} h_{a^{-1}bx}\bigl(d\alpha_{a^{-1},bx}(d\alpha_{b,x}(u)),d\alpha_{a^{-1},bx}(d\alpha_{b,x}(v))\bigr)\,d\mu_{T_1}(a). \end{align*} Because $T_1$ is abelian, the composition $\alpha_{a^{-1}}\circ\alpha_b$ equals $\alpha_{a^{-1}b}$. Set $c=b^{-1}a$, so $a^{-1}b=c^{-1}$. Haar measure is invariant under left translation, so $d\mu_{T_1}(a)=d\mu_{T_1}(c)$ under this substitution. Therefore the last integral becomes \begin{align*} \int_{T_1} h_{c^{-1}x}\bigl(d\alpha_{c^{-1},x}(u),d\alpha_{c^{-1},x}(v)\bigr)\,d\mu_{T_1}(c)=\bar h_x(u,v). \end{align*} Thus every element of $T_1$ acts by an isometry of $(M,\bar h)$. [/guided] [/step] [step:Apply the torus fixed point theorem to obtain a fixed coset] Write $\chi(N)$ for the Euler characteristic of a compact manifold $N$. We use the following standard fixed point input, not yet linked to a theorem in the wiki: if a compact torus acts isometrically on a compact connected Riemannian manifold whose Euler characteristic is nonzero, then the action has a fixed point. We also use the standard Euler characteristic input for maximal tori, not yet linked to a theorem in the wiki: if $G$ is compact and connected and $T\le G$ is a maximal torus, then \begin{align*} \chi(G/T)\ne 0. \end{align*} The preceding steps verify that $M=G/T_2$ is compact and connected and that $T_1$ acts isometrically on $M$ with respect to $\bar h$. Since $T_2$ is a maximal torus, the Euler characteristic input gives \begin{align*} \chi(M)=\chi(G/T_2)\ne 0. \end{align*} The fixed point theorem therefore gives a point $x_0\in M$ such that \begin{align*} t\cdot x_0=x_0 \end{align*} for every $t\in T_1$. Since every point of $M$ is a left coset, choose $g\in G$ such that $x_0=gT_2$. Then \begin{align*} tgT_2=gT_2 \end{align*} for every $t\in T_1$. [/step] [step:Translate fixedness into an inclusion of tori] From \begin{align*} tgT_2=gT_2 \end{align*} we obtain \begin{align*} g^{-1}tgT_2=T_2 \end{align*} for every $t\in T_1$, hence $g^{-1}tg\in T_2$ for every $t\in T_1$. Therefore \begin{align*} g^{-1}T_1g\le T_2. \end{align*} The subgroup $g^{-1}T_1g$ is a torus, because conjugation by $g$ is a Lie [group isomorphism](/page/Group%20Isomorphism) \begin{align*} T_1&\to g^{-1}T_1g \end{align*} \begin{align*} t&\mapsto g^{-1}tg, \end{align*} and it preserves compactness, connectedness, and abelianness. [/step] [step:Use maximality to upgrade the inclusion to equality] By definition, a maximal torus is a torus maximal among connected compact abelian Lie subgroups of $G$. Since $T_1$ is maximal, its conjugate $g^{-1}T_1g$ is also a maximal torus of $G$. The inclusion \begin{align*} g^{-1}T_1g\le T_2 \end{align*} places one maximal torus inside another torus of $G$, so maximality forces \begin{align*} g^{-1}T_1g=T_2. \end{align*} Conjugating this equality by $g$ gives \begin{align*} T_1=gT_2g^{-1}. \end{align*} Equivalently, replacing $g$ by $g^{-1}$, there exists an element of $G$ satisfying \begin{align*} gT_1g^{-1}=T_2. \end{align*} This is the desired conjugacy of maximal tori. [/step]