why might this demonstration regarding subgroups not be simple and obvious? If $H$ is a subgroup from $G$ so why:
$$a,b\in H\implies ab^{-1}\in H.$$
I thought that if $b\in H$ so $b^{-1} \in H$ since $H$ is a group as well, as $H$ is a subgroup so that's closed under the operator $(\cdot)$ from $G$ ($ab=a\cdot b$) so $ab \in H$ but because $b^{-1} \in H$ so $ab^{-1}\in H$.
But my professor put another demonstration out harder than this which I can't even understand. He used the fact that the null element and inverse for both $G$ and $H$ are the same.
Is my demonstration wrong?
 A: Your prof is establishing some basic results which you are using  in your proof: That the identity of $H$ is the identity of $G,$ and that if $b\in H$ then the inverse of  $b$ in the group $H$ is equal to the inverse of $b$ in the group $G.$
Let $1_G$ be the identity of $G$. Let $1_H$ be the identity of $H$.
(1a). For any $x,x'\in G$ we have $xx'=x$ iff $x'=1_G$...(1b).  In the case $x=x'=1_H$, since $1_H$ is the identity of $H$ and since $x,x'\in H$, we have $xx'=x1_H=x$ , so we conclude by (1a) that $1_G=x'=1_H$.
(2a). For any $b\in G$ there is a unique $z\in G$ such that $bz=1_G.$ This $z$ is denoted as $b^{-1}.$... (2b). Any $b\in H$ has a unique inverse $b'\in H,$ that is, a unique $b'\in H$ such that  $bb'=1_H.$ By (1b) we have $1_G=1_H$, so $bb'=1_G,$ so by  (2a) we have $b^{-1}=b'\in H$.
(3). If $a,b\in H$ then $a,b^{-1}\in H$ by (2b), and the product of members of $H$ is in $H$, so $ab^{-1}\in H.$
Addendum. An example of a structure $G$ that is $not$ a group, and a sub-structure $H$ where none of this works. Let $G$ be the set of all functions from $\Bbb R$ to $\Bbb R.$ For $f,g\in G$, define the function $fg\in G$ by $(fg)(r)=f(r)g(r)$ for each $r\in\Bbb R.$ The function $1_G,$ where $1_G(r)=1$ for every $r\in \Bbb R,$ is the only $f\in G$ such that $fg=g$ for $all$ $g\in G$... Let $H=\{f\in G: f(0)=0\}.$ Then $H$ has an identity too: $1_H(r)=1$ if $0\ne r\in \Bbb R,$ and $1_H(0)=0.$
