Proof
(related to Proposition: Criteria for Subgroups)
Proof for criterion $(1)$
"\(\Rightarrow\)"
 Let \(H\subseteq G\) be a subgroup of the group \((G,\ast)\).
 From the definition of the subgroup we have for $a,b\in H$ that $a\ast b^{1}\in H.$
"\(\Leftarrow\)"
 Assume \(a\ast b^{1}\in H\) for all \(a,b\in H\).
 If \(a=b\), we have \(b\ast b^{1}=e\), which means $e\in H$, i.e. $H$ contains the neutral element.
 If \(a=e\), we have \(e\ast b^{1}=b^{1}\), which means if $b\in H$ then $b^{1}\in H$, i.e. $H$ contains with $b$ also its unique inverse element $b^{1}.$
 If \(a,b\in H\), then according to the above result, also \(b^{1}\in H\).
 By assumption, $a\ast(b^{1})^{1}=a\ast b\in H$.
 Therefore $H$ is closed under the operation $"\ast".$
Proof for criterion $(2)$
 For any given \(a,b\in H\), we have \(a,b \in H_i\) for every \(i\in I\).
 Therefore also \(a^{1},b^{1}\in H_i\).
 By the definition of set intersection, also \(a^{1},b^{1}\in H\).
 In particular, \(a\ast b^{1}\in H\).
 By criterion $(1),$ \(H\) is a subgroup of $G.$
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References
Bibliography
 Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013