Do these numbers form a group under standard multiplication? $$e^{2πi(m/n)}$$ where $n \in N^{*}$ and $m=1,2,....n-1$
I think that there isn't an identity element cause $m$ is never $0$. Is that correct??
Do these numbers form a group? $e^{2πi(m/n)},m=1,2,....n-1,n \in N^{*}$
0
$\begingroup$
abstract-algebra
group-theory
finite-groups
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0Yes, that's right. – 2017-02-19
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0Another way to see it is: The set is not closed under multiplication. For example, if $a=e^{2πi(1/n)}$ and $b=e^{2πi((n-1)/n)}$, then $ab$ is not in the set. – 2017-02-19
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0@quasi $m=n$ gives the identity – 2017-02-19
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0@quasi It's a little unclear because there are two commas, but $m$ is allowed to be $n$ apparently, in which case it is a group. – 2017-02-19
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0I interpreted it as $m$ going from $1$ to $n-1$. Look at the title of the posted question. – 2017-02-19
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0@quasi I think you're right. the last $n$ is written to tell us that $n$ is a natural number. It's not part of delimiters for $m$. IMO. – 2017-02-19
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0@figarozo: Bottom line: If $m$ goes from $1$ to $n$, then it's a group. If $m$ goes from $1$ to $n-1$, then it's not a group. But now you can see how a minor notational confusion can wreck things! – 2017-02-19
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0thank you all guys yes m cannot be n only n-1 – 2017-02-19
1 Answers
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You are correct. It is not a group because there is no identity.
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0but m is only 1,2,3,...n-1????not n – 2017-02-19
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0@figarozo You wrote $1,2,\ldots.n-1,,n$, to be precise. So the $n$ is there. You're saying it's a typo? – 2017-02-19
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0yes yes sorry my fault then i ll fix it – 2017-02-19
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0i hope that's more clear – 2017-02-19
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0@figarozo Yes. I've edited my answer accordingly. – 2017-02-19