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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

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013
2. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013