# 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