I have a question while reading the following proof on William M.Boothby's textbook :
Thm3.6. An oriented, compact, conncted Lie group G has a unique bi-invariant volume element $\Omega$ such that $Vol G =1$.
The question I want to ask is the existence part. Since the author has already showed that every Lie group has a left-invariant volume element (as the $Corollary ~3.5) $, he verifies that $\Omega$ is right invariant (volume element). Thus, he shows that $Ad(g)^{*} \Omega_{e}= \Omega_{e}$. When considering $Ad:G \to gl(\mathfrak{g}), g \mapsto \sum_{i=1}^{n} \alpha_{ij}(g)X_i$, and $g \mapsto (\alpha_{ij}(g))$ defines a $C^{\infty}$-homomorphism of $G \to Gl(n,\mathbb{R})$ Thus, $Ad(g)^{*} \Omega_{e}= (det(\alpha_{ij}))^{1/2}\Omega_{e}$. But I do not understand this part on the textbook:
However, Since G is compact and connceted, the same applies to its image under the $C^{\infty}$-homomorphism of $G \to \mathbb{R}^{*}$,the multiplicative group of nonzero real numbers. However, the only compact connceted subgroup of $\mathbb{R}^{*}$ is $+1$...... (*)
Obviously, since (*) holds, in other words, $(det(\alpha_{ij}))^{1/2}=+1$ the existence of bi-invariant volume element is verified, but I don't know why the value of the determinant is only $+1$ just because of a compact and connected Lie group G.
Suppose $H\subset\Bbb R^*$ is a compact, connected subgroup. Then $H$ is a closed interval $[a,b]\subset\Bbb R_+$. Since $H^2\subset H$, we have $a^2\ge a$ and $b^2\le b$, so $b\le 1\le a$, which says that $a=b=1$ and $H=\{1\}$.