Why the structure group of principal $G$-bundle must be subgroup of $G$?

Question

As picture below, seemly, it is said that the structure group of principal $G$-bundle must be subgroup of $G$. Why ?

Picture below is from the 65 page of Jost's Riemannian Geometry and Geometric Analysis .

Answer 1

Cosider two locally trivial $$ \varphi_{_U} :\pi^{-1}(U)\rightarrow U\times G \\ \varphi_{_V} :\pi^{-1}(V)\rightarrow V\times G $$ So, $$ \varphi_{_V} \circ \varphi^{-1}_{_U} (x,g) \rightarrow (x,\varphi_{_{VU}}(x)g) $$ because $\varphi_{_{VU}}(x)g\in G$, assume $\varphi_{_{VU}}(x)g=h$ , then $\varphi_{_{VU}}(x)=hg^{-1}\in G$ . So, the structure group is subgroup of $G$.