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I'm stuck on the following exercise from Herstein's "Topics in Algebra":

"Show that if $(a,n)=1$ then you can find $[b]\in J_n$ such that $[a][b]=1$"

(note: $[a]$ is the congruence class (mod $n$) of $a$, and $J_n$ is the set of congruence classes mod $n$)

I've been thinking about using the property $(a,n)=1\Rightarrow \exists h,k\in\mathbb{Z}$ such that $ha+kn=1$ and also the Euclidean algorithm by writing $a=qn+b$ $0\leq b<a$ (since $J_n=\{[0],...[n-1]\})$ but I haven't been able to come up with a proof.

So, I would appreciate any comment/hint about how to prove this fact.

Best regards,

lorenzo

lorenzo
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    With this notation, $[a][b]=[1]$. – ajotatxe Mar 04 '17 at 12:16
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    So you know that there exist $h, k$ such that $ha + kn = 1$. What if you take $b = h$? – Kenny Wong Mar 04 '17 at 12:21
  • @KennyWong that's exactly the point where I got stuck – lorenzo Mar 04 '17 at 12:25
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    What is the class of $kn$ modulo $n$? – Kenny Wong Mar 04 '17 at 12:26
  • @KennyWong I'd say it is $n$ (since it is $n|(kn-n)$) but I'm confused by the fact that the congruence classes go from $[0]$ to $[n-1]$; I'm new to the concept of congruence and I think I'm getting something wrong. – lorenzo Mar 04 '17 at 12:38
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    Let's use actual numbers. What is the class of $3$ modulo $3$? (The options are $[0]$, $[1]$ or $[2]$.) – Kenny Wong Mar 04 '17 at 12:39
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    In fact, let me rephrase the question: what remainder do you get when you divide $3$ by $3$? – Kenny Wong Mar 04 '17 at 12:41
  • @KennyWong you get a remainder of 0 and so (if that's the meaning of class) the class of 3 modulo 3 is [0] and by the same reasoning the class of $kn$ modulo $n$ is also $[0]$; is this correct? – lorenzo Mar 04 '17 at 12:50
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    That's absolutely right! Now, what is the class of $1 - kn$ modulo $n$? – Kenny Wong Mar 04 '17 at 12:51
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    What is the class of $-kn$, first of all? For example, what is the class of $-23*3$ modulo $3$? – Kenny Wong Mar 04 '17 at 13:03
  • @KennyWong I gave the previous answer considering $1-kn$ mod $a$ sorry; the class of $kn$ mod $n$ and also of $-23*3$ mod $3$ should also be $[0]$ and since $(1-kn)/n =1/n -k$ I think the class of $1-kn$ mod $n$ should be $[1]$ am I right? – lorenzo Mar 04 '17 at 13:13
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    Yes. So the class of $ha$ is... – Kenny Wong Mar 04 '17 at 13:17
  • @KennyWong is also $[1]$ also so if we take $b=h$ we have $[ab]=[a][b]=1$, as desired. Thank you very much. If you write your comments as an answer I'll accept it. – lorenzo Mar 04 '17 at 13:25
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    Why don't you write it as an answer yourself? I'll be happy to accept it. :) – Kenny Wong Mar 04 '17 at 13:47

1 Answers1

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(note: this answer is a summary of Kenny Wong's comments to my question above)

$(a,n)=1\Rightarrow \exists h,k\in\mathbb{Z}$ such that $ha+kn=1\Rightarrow ha=1-kn$ and since $[1-kn]=[1]$ we have $[ha]=[1]$ so if we take $b=h$ we get $[ab]=[a][b]=[1]$, as desired.

lorenzo
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