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Let $G$ be a group. Show that a nonempty subset $H$ is a subgroup of $G$ if any only if $ab^{-1} \in H $ for any $a,b \in H$.

The forward direction is quite easy. Suppose $H$ is a subgroup. Then by closure, $ab \in H$ for any $a,b \in H$. Every element has an inverse. Hence, if $b \in H $, then $b^{-1} \in H$. Hence, by closure again, $ab^{-1} \in H$.

Backward direction, suppose $ab^{-1} \in H $ for any $a,b \in H$. Let $a=b$. Then we have $bb^{-1}=1_H\in H$. Let $a=1_H$ , we have $1_Hb^{-1}=b^{-1} \in H$. I don't know how to show closure.

Reference: Fraleigh p. 58 Question 5.45 in A First Course in Abstract Algebra

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$ab=a(b^{-1})^{-1}$. –  user1551 Jan 31 '13 at 15:40
@JasonBourne thanks. I have edited. –  Idonknow Jan 31 '13 at 15:48

2 Answers 2

up vote 4 down vote accepted

Take $a$, $b$ in $H$ then $c=b^{-1}\in H$, so that $a c^{-1}=a (b^{-1})^{-1}=ab\in H$.

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@Quimey Can you please flesh out why you know $b^{-1} \in H$? For the backward direction, we only know $H$ is a nonempty subset. It doesn't have to be a group. Hence it doesn't have to contain inverses?

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You should put a comment under @Quimey's answer to ask him your question. The OP already know why $b^{-1} \in H$, see his question. –  scaaahu Dec 25 '13 at 9:32

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