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The subgroup {1,2,9} contains the identity element, also, each of its element has its inverse but how about closure? Is it correct to deduce closure from the fact that $2 \times_{17} 9 = 1$?

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    $\begingroup$ What about $2\times_{17}2$? $\endgroup$
    – Arthur
    Commented May 4, 2016 at 12:31
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    $\begingroup$ $2\times_{17}2=4$ and $4$ is not in {1,2,9} and therefore, not closed! Forgot that one, thanks a lot. $\endgroup$
    – ZeroCool
    Commented May 4, 2016 at 12:35

2 Answers 2

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Hint:If $H$ is a subgroup of $G$ then $O (H)$ divides $O (G)$

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The subset {1,2,9} is not closed under $\times_{17}$ because $2\times_{17}2=4$ and $4$ is not in {1,2,9}.

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