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January 29th, 2017, 11:23 AM   #1
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H subset of a group G. H subgroup iff H not empty and g,h in H then $gh^{-1}$ is in H


I have this proposition, I write all proof but, there is a part that I don't understand.

Let $H$ be a subset of a group $G$. Then $H$ is a subgroup of $G$ if and only if $H \ne \emptyset$, and whenever $g,h \in H$ then $gh^{-1} \in H$.

First, assume that $H$ is a subgroup of $G$. We want to show that $gh^{-1} \in H$ whenever $g,h \in H$.
Since $h \in H$, its inverse $h^{-1} \in H$. Because of the closure of the group operation, $gh^{-1} \in H$.
Conversely, suppose $H \subset G \, | \, H \ne \emptyset, gh^{-1} \in H$ whenever $g,h \in H$.
If $g \in H$, then $gg^{-1} = e \in H$.
If $g \in H$, then $eg^{-1} = g^{-1} \in H$.
Now let $h_1, h_2 \in H$. We must show that their product is also in $H$.
However $h_1(h_2^{-1})^{-1} = h_1h_2 \in H$. Hence, $H$ is a subgroup of $G$.

In the first part $(\Rightarrow)$ I think it's clear that:
Since $h \in H$, its inverse $h^{-1} \in H$. Because of the closure of the group operation, $gh^{-1} \in H$.
because we are assuming that $H$ is a subgroup of $G$.

But, in the second part $(\Leftarrow)$, there is:
If $g \in H$, then $gg^{-1} = e \in H$.
I don't understand. Since we are NOT assuming $H$ as a subgroup of $G$, why $g^{-1}$ would be also $\in H$?

Please, can you help me? Many thanks!
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January 29th, 2017, 06:35 PM   #2
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in the $\Leftarrow$ direction we assume that $\forall g,~h \in H,~gh^{-1}\in H$

just let $h=g$

$g\in H \Rightarrow g g^{-1} \in H$
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