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If $n$ is a positive number, let $\mathbb{Z}_n^*$ be the subset of class representatives in $\{1,..., n-1\}$ which are relatively prime to $n$. Show that if $[a],[b] \in \mathbb{Z}_n^* $ then $[ab] \in \mathbb{Z}_n^*$.

I think $[a]$ is an equivalence class. But what does it represent? If $a = 1$, what does $[1]$ imply?

Update: this is what I understand so far. For example, let n = 5, a = 2. Then relatively primes of n is {1,2,3,4}, and [a] = {2,4} because it's only 2 in {1,2,3,4} that has residual when divided by n = 5. Is this correct?

Update: My thought on proving the problem. Let [a] = {a + kn | k $\in \mathbb{Z}_n^*$}, [b] = {b + ln | l $\in \mathbb{Z}_n^*$}. Pick an a' from [a] and b' from [b]. I can say a' = a + kn, b' = b +ln. Then a'*b' = ab + n(ak + bk + $nk^2$). This is congruent to ab mod n. Therefore [ab] $\in \mathbb{Z}_n^*$. Would this be an appropriate proof?

jun
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    It is an equivalence class on $\mathbb{Z}$: integers $a,b$ are equivalent if $a-b$ is divisible by $n$. Now, $\mathbb{Z_n}^*$ is the collection of the equivalence classes $[a]$ where $a$ is relatively prime to $n$. (this does not depend on the choice of the representatives) – Mark Aug 31 '20 at 12:29
  • so if a =1, then [a] = {1} ? sorry. I still don't get it. Can you give me an example, please – jun Aug 31 '20 at 12:33
  • um. "integers a,b are equivalent if a−b is divisible by n" is this line true by default? I thought the meaning of equivalency depends on how we define it. – jun Aug 31 '20 at 12:41
  • The equivalence relation here depends on $n$. It is not “true by default” but “true by the meaning of what your notation means”. What do you think the notation $[a]$ used in your question is supposed to mean? Have you ever worked with some actual examples of elements of $\mathbf Z_n^*$? – KCd Aug 31 '20 at 12:44
  • I thought Zn* is just a symbol. Does it have its own meaning? I think [a] as I wrote in my question. I updated it. – jun Aug 31 '20 at 12:50
  • For example, if $n=2$ then $[1]={...,-5,-3,-1,1,3,5,7,9,11,...}$. In this case $[1]$ is the set of all odd numbers. – Mark Aug 31 '20 at 12:51
  • The question is ill posed: it first defined $\mathbb Z_n^$ as a set of representatives, so the elements are integers. Then it states “$[a]\in \mathbb Z_n^$” implying that the elements of $\mathbb Z_n^*$ are now equivalence classes. Which one is it? – Christoph Aug 31 '20 at 12:56
  • Semantics aside... this is just asking you to show that if $a$ is relatively prime to $n$ and $b$ is also relatively prime to $n$ then $a\times b$ is also relatively prime to $n$. Further, if any number $A$ has the same remainder as $a$ when divided by $n$ and any number $B$ has the same remainder as $b$ when divided by $n$ then $A\times B$ has the same remainder as $a\times b$ when divided by $n$. – JMoravitz Aug 31 '20 at 13:02
  • You are getting bogged down by notation which wasn't supposed to be the challenge of this problem but still a necessary thing to get used to. With $n=5$, what was intended was for $[2]$ to represent the set of all integers who have remainder $2$ when divided by $n$ which happens to be $[2]={\dots,-8,-3,2,7,12,\dots}$. Now... when $n$ happened to be prime you'll have $1,2,\dots,n-1$ are all coprime to $n$ and this is a particularly important example you'll revisit later. More importantly for now however is when $n$ is not prime, we'll be skipping some of those numbers yet it remains closed – JMoravitz Aug 31 '20 at 13:08
  • In regards to your update, you have shown that $[a]\cdot [b]$ is well defined, but you haven't yet shown that $[ab]\in\Bbb Z_n^*$ specifically, only that $[ab]\in\Bbb Z_n$. You still need to show that $ab$ is coprime to $n$ so long as $a$ and $b$ are coprime to $n$. – JMoravitz Aug 31 '20 at 14:41
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    The titled inference is a frequent DUPE, e.g. see How to show $a,b$ coprime to $n\Rightarrow ab$ coprime to $n$? – Bill Dubuque Aug 31 '20 at 14:41

1 Answers1

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In $\mathbb{Z}$, we define an equivalence relation by : $$ a \text{ and } b \text{ are equivalent } \iff n \mid b-a $$ so $[a]=\{ b\in \mathbb{Z} : a \text{ and } b \text{ are equivalent }\}=\{ b \in \mathbb{Z} : n|b-a\}=\{ b\in \mathbb{Z} : b-a=nk \text{ s.t } k\in \mathbb{Z}\}=\{a+nk : k\in \mathbb{Z}\},$

Thus in $\mathbb{Z}_3$, for example $[1]=\{1+3k : k\in \mathbb{Z}\}=\{\cdots,-2,1,4,7,\cdots \}.$

H_K
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  • I have edited my answer. Thank you @halrankard2. – H_K Aug 31 '20 at 13:15
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    Looks good. You might add explicitly that $\mathbb{Z}_n$ is defined to be the set of equivalence classes. That would help in reconciling the example with $n=3$. – halrankard2 Aug 31 '20 at 13:20