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January 29th, 2017, 11:23 AM  #1  
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7  H subset of a group G. H subgroup iff H not empty and g,h in H then $gh^{1}$ is in H
Hi, I have this proposition, I write all proof but, there is a part that I don't understand. Proposition: 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 $gh^{1} \in H$. Proof: $(\Rightarrow)$ 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, $gh^{1} \in H$. $(\Leftarrow)$ 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 $G$. $\Box$ In the first part $(\Rightarrow)$ I think it's clear that: Quote:
But, in the second part $(\Leftarrow)$, there is: Quote:
Please, can you help me? Many thanks!  
January 29th, 2017, 06:35 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,426 Thanks: 1314 
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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$gh1$, empty, group, iff, subgroup, subset 
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