[proofplan] We prove the subgroup statement directly from the time-one-map definition of $\operatorname{Ham}(M,\omega)$. First, a Hamiltonian isotopy preserves $\omega$ by Cartan's formula and hence gives a path in $\operatorname{Symp}_0(M,\omega)$. The zero Hamiltonian gives the identity. Finally, differentiating the composed path and the inverse path gives explicit Hamiltonians for composition and inverse; the same formulas preserve the compact-support convention on noncompact manifolds. [/proofplan] [step:Show that every Hamiltonian time-one map lies in $\operatorname{Symp}_0(M,\omega)$] Let \begin{align*} H:[0,1]\times M\to\mathbb R \end{align*} be a smooth time-dependent Hamiltonian, compactly supported in $M$ uniformly in $t$ when $M$ is noncompact. Let $\mathfrak X(M)$ denote the space of smooth vector fields on $M$. For each $t\in[0,1]$, let $H_t:M\to\mathbb R$ denote the smooth function $H_t(x)=H(t,x)$, and let $X_t\in\mathfrak X(M)$ be the time-dependent Hamiltonian vector field defined by \begin{align*} \iota_{X_t}\omega=dH_t. \end{align*} Let $\varphi_t:M\to M$ be the Hamiltonian flow, defined by \begin{align*} \frac{d}{dt}\varphi_t(x)=X_t(\varphi_t(x)) \end{align*} and $\varphi_0=\operatorname{id}_M$. In the noncompact case, the uniform compact-support hypothesis makes $X_t$ compactly supported in a fixed compact set, so the flow exists on the full interval $[0,1]$. We compute the derivative of the pulled-back symplectic form along the isotopy. Since $d\omega=0$ and $\iota_{X_t}\omega=dH_t$, Cartan's formula gives \begin{align*} \mathcal L_{X_t}\omega=d(\iota_{X_t}\omega)+\iota_{X_t}(d\omega)=d(dH_t)+0=0. \end{align*} Therefore \begin{align*} \frac{d}{dt}\varphi_t^*\omega=\varphi_t^*(\mathcal L_{X_t}\omega)=0. \end{align*} Since $\varphi_0^*\omega=\omega$, it follows that \begin{align*} \varphi_t^*\omega=\omega \end{align*} for every $t\in[0,1]$. Thus $\varphi_t\in\operatorname{Symp}(M,\omega)$ for every $t$, and the path $t\mapsto\varphi_t$ connects $\operatorname{id}_M$ to $\varphi_1$ inside $\operatorname{Symp}(M,\omega)$. Hence $\varphi_1\in\operatorname{Symp}_0(M,\omega)$. [guided] Let \begin{align*} H:[0,1]\times M\to\mathbb R \end{align*} be a smooth time-dependent Hamiltonian. When $M$ is noncompact, the convention in the statement says that the supports of the functions $H_t:M\to\mathbb R$, defined by $H_t(x)=H(t,x)$, are all contained in one compact subset of $M$. This matters because the associated Hamiltonian vector fields are then compactly supported in a fixed compact set, which ensures that the flow exists for all $t\in[0,1]$. For each time $t$, define the Hamiltonian vector field $X_t\in\mathfrak X(M)$ by the sign convention \begin{align*} \iota_{X_t}\omega=dH_t. \end{align*} Let $\varphi_t:M\to M$ be the flow of $X_t$, so \begin{align*} \frac{d}{dt}\varphi_t(x)=X_t(\varphi_t(x)) \end{align*} and $\varphi_0=\operatorname{id}_M$. We want to show that each $\varphi_t$ preserves the symplectic form. The standard way to check this is to differentiate $\varphi_t^*\omega$ with respect to $t$. The derivative formula for pullbacks along a flow gives \begin{align*} \frac{d}{dt}\varphi_t^*\omega=\varphi_t^*(\mathcal L_{X_t}\omega). \end{align*} Now Cartan's formula expresses the Lie derivative of a differential form as \begin{align*} \mathcal L_{X_t}\omega=d(\iota_{X_t}\omega)+\iota_{X_t}(d\omega). \end{align*} The first term is $d(dH_t)$ because $\iota_{X_t}\omega=dH_t$, and the second term vanishes because $\omega$ is closed. Hence \begin{align*} \mathcal L_{X_t}\omega=d(dH_t)+\iota_{X_t}(0)=0. \end{align*} Therefore \begin{align*} \frac{d}{dt}\varphi_t^*\omega=0. \end{align*} So $\varphi_t^*\omega$ is independent of $t$. Evaluating at $t=0$ gives \begin{align*} \varphi_t^*\omega=\varphi_0^*\omega=\operatorname{id}_M^*\omega=\omega. \end{align*} Thus every $\varphi_t$ is a symplectomorphism. Since the path $t\mapsto\varphi_t$ begins at $\operatorname{id}_M$ and ends at $\varphi_1$, the time-one map $\varphi_1$ lies in the identity component $\operatorname{Symp}_0(M,\omega)$. [/guided] [/step] [step:Generate the identity by the zero Hamiltonian] Define the zero Hamiltonian $0:[0,1]\times M\to\mathbb R$ by $0(t,x)=0$. Its Hamiltonian vector field is the zero vector field, since $\iota_{0}\omega=0=d0_t$. The corresponding flow is $\varphi_t=\operatorname{id}_M$ for every $t\in[0,1]$. Hence $\operatorname{id}_M\in\operatorname{Ham}(M,\omega)$. [/step] [step:Compute a Hamiltonian for the composition of two Hamiltonian isotopies] Let $H,K:[0,1]\times M\to\mathbb R$ be smooth time-dependent Hamiltonians satisfying the compact-support convention when $M$ is noncompact. Let $X_t,Y_t\in\mathfrak X(M)$ be the corresponding Hamiltonian vector fields, so $\iota_{X_t}\omega=dH_t$ and $\iota_{Y_t}\omega=dK_t$. Let $\varphi_t:M\to M$ and $\psi_t:M\to M$ be their flows. Define $\theta_t:M\to M$ by $\theta_t=\varphi_t\circ\psi_t$. We claim that $\theta_t$ is Hamiltonian with Hamiltonian $L:[0,1]\times M\to\mathbb R$ given by $L_t=H_t+K_t\circ\varphi_t^{-1}$. For $x\in M$, differentiation gives $\frac{d}{dt}\theta_t(x)=X_t(\theta_t(x))+d(\varphi_t)_{\psi_t(x)}(Y_t(\psi_t(x)))$. Thus the time-dependent generating vector field $Z_t\in\mathfrak X(M)$ of $\theta_t$ is $Z_t=X_t+(\varphi_t)_*Y_t$. Because $\varphi_t$ is symplectic, for every $z\in M$ and every $v\in T_zM$, $\omega_z((\varphi_t)_*Y_t(z),v)=\omega_{\varphi_t^{-1}(z)}(Y_t(\varphi_t^{-1}(z)),d(\varphi_t^{-1})_z(v))$. Using $\iota_{Y_t}\omega=dK_t$, the right-hand side is $dK_t|_{\varphi_t^{-1}(z)}(d(\varphi_t^{-1})_z(v))=d(K_t\circ\varphi_t^{-1})_z(v)$. Therefore $\iota_{(\varphi_t)_*Y_t}\omega=d(K_t\circ\varphi_t^{-1})$. Together with $\iota_{X_t}\omega=dH_t$, this gives $\iota_{Z_t}\omega=dH_t+d(K_t\circ\varphi_t^{-1})=dL_t$. Hence $\theta_t$ is a Hamiltonian isotopy and $\theta_1=\varphi_1\circ\psi_1$ lies in $\operatorname{Ham}(M,\omega)$. If $M$ is noncompact and the supports of $H_t$ and $K_t$ are contained in compact sets $C_H,C_K\subset M$ independent of $t$, then $\operatorname{supp}L_t\subset C_H\cup\varphi_t(C_K)$. The set $\bigcup_{t\in[0,1]}\varphi_t(C_K)$ is compact as the image of the compact set $[0,1]\times C_K$ under the smooth map $(t,x)\mapsto\varphi_t(x)$. Thus $L$ satisfies the same compact-support convention. [/step] [step:Compute a Hamiltonian for the inverse of a Hamiltonian isotopy] Let $H:[0,1]\times M\to\mathbb R$ be a smooth time-dependent Hamiltonian satisfying the compact-support convention when $M$ is noncompact. Let $X_t\in\mathfrak X(M)$ be defined by $\iota_{X_t}\omega=dH_t$, and let $\varphi_t:M\to M$ be its flow. Define \begin{align*} \eta_t:M\to M \end{align*} by $\eta_t=\varphi_t^{-1}$. We claim that $\eta_t$ is Hamiltonian with Hamiltonian \begin{align*} G:[0,1]\times M\to\mathbb R,\qquad G_t=-H_t\circ\varphi_t. \end{align*} The identity $\eta_t\circ\varphi_t=\operatorname{id}_M$ gives, after differentiating at a point $x\in M$, \begin{align*} \frac{d}{dt}\eta_t(\varphi_t(x))+d(\eta_t)_{\varphi_t(x)}(X_t(\varphi_t(x)))=0. \end{align*} Hence the generating vector field $W_t\in\mathfrak X(M)$ of $\eta_t$ satisfies \begin{align*} W_t(x)=-d(\varphi_t^{-1})_{\varphi_t(x)}(X_t(\varphi_t(x))). \end{align*} For $x\in M$ and $v\in T_xM$, using that $\varphi_t$ is symplectic gives \begin{align*} \omega_x(W_t(x),v)=-\omega_{\varphi_t(x)}(X_t(\varphi_t(x)),d(\varphi_t)_x(v)). \end{align*} Since $\iota_{X_t}\omega=dH_t$, the right-hand side equals \begin{align*} -dH_t|_{\varphi_t(x)}(d(\varphi_t)_x(v))=d(-H_t\circ\varphi_t)_x(v). \end{align*} Thus \begin{align*} \iota_{W_t}\omega=dG_t. \end{align*} So $\eta_t$ is a Hamiltonian isotopy and $\eta_1=\varphi_1^{-1}$ lies in $\operatorname{Ham}(M,\omega)$. If $M$ is noncompact and $\operatorname{supp}H_t\subset C_H$ for a compact set $C_H\subset M$ independent of $t$, then \begin{align*} \operatorname{supp}G_t\subset\varphi_t^{-1}(C_H). \end{align*} The set $\bigcup_{t\in[0,1]}\varphi_t^{-1}(C_H)$ is compact as the image of $[0,1]\times C_H$ under the smooth map $(t,x)\mapsto\varphi_t^{-1}(x)$. Hence $G$ also satisfies the compact-support convention. [guided] Let $H:[0,1]\times M\to\mathbb R$ be a smooth time-dependent Hamiltonian, and let $X_t\in\mathfrak X(M)$ be defined by \begin{align*} \iota_{X_t}\omega=dH_t. \end{align*} Let $\varphi_t:M\to M$ be its Hamiltonian flow. We need to show that the inverse path $\eta_t=\varphi_t^{-1}$ is again generated by a Hamiltonian vector field. The first question is: what is the vector field generating the inverse path? Since \begin{align*} \eta_t\circ\varphi_t=\operatorname{id}_M, \end{align*} differentiating this identity at a point $x\in M$ gives \begin{align*} \frac{d}{dt}\eta_t(\varphi_t(x))+d(\eta_t)_{\varphi_t(x)}(X_t(\varphi_t(x)))=0. \end{align*} Thus the velocity of the inverse path at the point $x$ is \begin{align*} W_t(x)=-d(\varphi_t^{-1})_{\varphi_t(x)}(X_t(\varphi_t(x))). \end{align*} This defines the time-dependent vector field $W_t\in\mathfrak X(M)$ generating $\eta_t$. Now we identify its Hamiltonian. For $x\in M$ and $v\in T_xM$, use that $\varphi_t$ is symplectic, so \begin{align*} \omega_x(d(\varphi_t^{-1})_{\varphi_t(x)}a,v)=\omega_{\varphi_t(x)}(a,d(\varphi_t)_x(v)) \end{align*} for every $a\in T_{\varphi_t(x)}M$. Applying this with $a=X_t(\varphi_t(x))$ and including the minus sign in the definition of $W_t$, we obtain \begin{align*} \omega_x(W_t(x),v)=-\omega_{\varphi_t(x)}(X_t(\varphi_t(x)),d(\varphi_t)_x(v)). \end{align*} Because $\iota_{X_t}\omega=dH_t$, the last expression is \begin{align*} -dH_t|_{\varphi_t(x)}(d(\varphi_t)_x(v)). \end{align*} By the chain rule, this equals \begin{align*} d(-H_t\circ\varphi_t)_x(v). \end{align*} Therefore \begin{align*} \iota_{W_t}\omega=d(-H_t\circ\varphi_t). \end{align*} So the inverse path is generated by the Hamiltonian $G_t=-H_t\circ\varphi_t$. Finally suppose $M$ is noncompact and $\operatorname{supp}H_t\subset C_H$ for a compact set $C_H\subset M$ independent of $t$. If $G_t(x)\ne0$, then $H_t(\varphi_t(x))\ne0$, so $\varphi_t(x)\in C_H$ and $x\in\varphi_t^{-1}(C_H)$. Hence \begin{align*} \operatorname{supp}G_t\subset\varphi_t^{-1}(C_H). \end{align*} The union of these supports over $t\in[0,1]$ is contained in the image of the compact set $[0,1]\times C_H$ under the smooth map $(t,y)\mapsto\varphi_t^{-1}(y)$. Therefore the inverse Hamiltonian also has uniformly compact support. [/guided] [/step] [step:Conclude the subgroup property] The preceding steps show that $\operatorname{Ham}(M,\omega)\subset\operatorname{Symp}_0(M,\omega)$, that $\operatorname{id}_M\in\operatorname{Ham}(M,\omega)$, that the composition of two elements of $\operatorname{Ham}(M,\omega)$ again lies in $\operatorname{Ham}(M,\omega)$, and that the inverse of every element of $\operatorname{Ham}(M,\omega)$ again lies in $\operatorname{Ham}(M,\omega)$. Therefore $\operatorname{Ham}(M,\omega)$ is a subgroup of $\operatorname{Symp}_0(M,\omega)$. [/step]