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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

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:
 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:
Quote:
 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! January 29th, 2017, 06:35 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 in the $\Leftarrow$ direction we assume that $\forall g,~h \in H,~gh^{-1}\in H$ just let $h=g$ $g\in H \Rightarrow g g^{-1} \in H$ Thanks from beesee Tags $gh1$, empty, group, iff, subgroup, subset Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post icemanfan Abstract Algebra 5 March 5th, 2012 03:35 PM johnmath Abstract Algebra 3 March 3rd, 2011 08:22 AM millwallcrazy Abstract Algebra 1 April 26th, 2010 07:18 AM envision Abstract Algebra 1 October 4th, 2009 03:24 AM sastra81 Abstract Algebra 1 January 3rd, 2007 07:58 AM

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