Is the set $S=\{\overline{1},\overline{3},\overline{4},\overline{11},\overline{14}\}\subseteq\mathbb{Z}_{14}$ an algebraic structure concerning multiplication of residue classes? For example, I can see that $\bar{4}\cdot\bar{4}=\overline{16}=\bar{2}\notin S$. Does this imply that $S$ is not an algebraic structure?
$\{\bar{1},\bar{3},\bar{4},\bar{11},\bar{14}\}\subseteq\mathbb{Z}_{14}$ an algebraic structure concerning multiplication of residue classes?
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linear-algebra
residue-calculus
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0You showed that $S$ is not closed with respect to multiplication. It is not closed with respect to addition either: $\overline{3} + \overline{4} = \overline{7} \notin S$ – 2017-02-17
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0If you are only concerned with multiplication then yes you don't need to worry about addition. I just thought I'd mention addition anyway. Sorry, didn't mean to throw you off – 2017-02-17
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0The answer below is spot on: your set isn't closed with respect to multiplication so isn't an algebraic structure concerning multiplication. The only reason I mentioned addition at all is because, if you wanted to, you could also inquire if it is an algebraic structure concerning addiition- and again the answer is no due to closure. But if you are not concerned with addition, you can ignore my comment. – 2017-02-17
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If nothing else, it just means that the set is not closed under multiplication. Closure is a very important property to have, especially for algebraic structures.