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I would like to verify my proof for the following exercise from Foot and Dummit's "Abstract Algebra": enter image description here

I will prove the following statement:

$$ a \in (Z/nZ)^* \iff gcd(a,n) = 1 $$

Thus, the set $(Z/nZ)^*$ is precisely characterized by the elements smaller than $n$ whose gcd is 1 and thus its size is $\phi(n)$.

Proof:

$ \rightarrow $ We assume by contradiction that there exists $J>1$ s.t. $gcd(a,n)=J$.

Let $a \in (Z/nZ)^*$, then there exists $b \in Z/nZ$ s.t. $ab = 1\,(mod\,n) $.

Thus, by definition, there exists $k \in Z$ s.t. $$ab-kn=1$$ Using the Euclidean algorith we compute $gcd(ab,n)$:

If we run the algorithm it should stop at the first iteration as $ab=kn+1 \rightarrow gcd(ab,n)=1$.

By our assumption $gcd(a,n)=J>1$ and in particular - $a|J$ and $n|J$. Contradition.

$\leftarrow$ Let $a \in Z$ s.t. $gcd(a,n)=1$. Then, by "Bezout's Identity" there exists $0<x,y \in Z$ s.t. $$ ax+ny=1 $$ Thus by definition: $$ ax = 1 (mod\,n) $$ And we have found $x = [b] \in Z/nZ$ as required. Meaning $ [a] \in (Z/nZ)^* $. $\square$

Let me know what you think?

Nimrodshn
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  • For a solution-verification question to be on topic you must specify which step in the proof you question, and why so. This site is not meant to be an open-ended proof checking machine. This common exercise is already solved here many times. You should compare your proof to those before posting. – Bill Dubuque Aug 06 '22 at 17:38
  • The proof is definitely on the right lines, but the Euclidean algorithm does not show that $ab - kn =1 \implies \text{gcd}(a,n)=1$. The result is true, but you need a different explanation. – legionwhale Aug 06 '22 at 17:40
  • @legionwhale Oh but it does show that $gcd(ab,n)=1$ which is enough for me :) this is simply by running the algorithm and stopping at the first iteration. – Nimrodshn Aug 06 '22 at 17:42
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    @Nimrodshn There is a much simpler and clearer explanation why the result is true. It's true that this step is part of the proof of the Euclidean algorithm. But this logical step is not the entire Euclidean algorithm, so it is a bit puzzling to refer to it as such. Your proof is otherwise concise and correct. – legionwhale Aug 06 '22 at 17:46
  • @legionwhale I guess you can say that its due to the "Division Algorithm"? Am not sure how to make the argument clearer. – Nimrodshn Aug 06 '22 at 17:51
  • $d|a$ and $d|n \implies d|ab-kn \implies d|1$. So $\text{gcd}(a,n)|1$. – legionwhale Aug 06 '22 at 20:52
  • @legionwhale Yup I learnt this argument after - I like it a lot, thanks! I still think that my argument is valid though. Even though slightly more "brute force", i.e lets run the "Euclidean algorithm" on "ab" and "n" and using the previous knowledge: $ab=kn+1 \rightarrow gcd(ab,n)=1$. – Nimrodshn Aug 07 '22 at 07:51
  • @Nimrodshn Sure; it's definitely not incorrect. It's good practice to always maximise clarity, though. – legionwhale Aug 07 '22 at 13:22

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