Show that (1) $\Longleftrightarrow$(2):
(1) For $H \subseteq G$ with $H \ne \varnothing$ of a group $(G,\ast)$, $(H,\ast)$ is a subgroup of $(G,\ast)$ if:
(G1): $\forall a,b \in H: a \ast b \in H$
(G2): $\forall a \in H: a^{-1} \in H$
(2) For $H \subseteq G$ with $H \ne \varnothing$ of a group $(G,\ast)$, $(H,\ast)$ is a subgroup of $(G,\ast)$ if:
(U): $\forall a,b \in H: a \ast b^{-1} \in H$
(Note: that $\ast$ is associative follows from $(G, \ast) $ being a group)
(1) $\Longrightarrow$ (2)
(G1) states $\forall a,b \in H: a \ast b \in H$
but since (G2) says $\forall a \in H: a^{-1} \in H$
$(\forall a,b \in H: a \ast b \in H) \overbrace{\Longleftrightarrow}^{(G2)} (\forall a,b \in H: a \ast b^{-1} \in H)$
(2) $\Longrightarrow$ (1)
We choose $a \in H$ and $a \in H$ another time. For (U) this means since $a$ is definitly a element of $H$,
that $a,a \in H: a \ast a^{-1} \in H$
But $a \ast a^{-1}= e$ by definition of the inverse. So $e \in H$ can be concluded by (U).
So since $e \in H$ we now choose any $a \in H$ and as second element $e \in H$.
(U) states $e,a \in H: e \ast a^{-1} \in H$
But $e \ast a^{-1}=a^{-1}$, so (U) gives us $e,a \in H: a^{-1} \in H$ which is (G2) since $e \in H$ no matter what.
But since for all elements in $H$ its inverse exists:
$(\forall a,b \in H: a \ast b^{-1} \in H) \overbrace{\Longleftrightarrow}^{(G2)} (\forall a,b \in H: a \ast b \in H)$ which is (G1)
$\Box$
Would be great if someone could look over it and give me some feedback if this is correct :)
I think you could improve a bit the implication $(U)\implies (G1)$. So you know that $a\star b^{-1}\in H$ for all $a,b\in H$ and you want to show that $H$ is closed under the group operation, that is, $a\star b\in H$ for all $a,b\in H$. To do this, just note that you may express the product $a\star b$ as $a\star (b^{-1})^{-1}$.