I tried to find an approach but no luck so far
Showing that $U(n) = \{m : 1 \le m < n, \gcd(m, n) = 1\}$ is a group under multiplication modulo $n$
0
$\begingroup$
group-theory
finite-groups
modular-arithmetic
-
0What do you mean, "find an approach"? Have you tried to show that $U(n)$ is closed under multiplication modulo $n$, that there are inverses, etc? – 2017-01-17
1 Answers
0
$m,m'$ in $U(n)$ implies there exists $a,b,a',b'$ such that $am+bn=1, a'm'+b'n=1$. This implies that $(am+bn)(a'm'+b'n)=aa'mm'+(amb'+a'm'b+bb'n)n=1$, we deduce that $gcd(mm',1)=1$. We deduce that $U(n)$ is stable by multiplication, the inverse of $m$ mod $n$ is $a$ such that $am+bn=1$.
If $n\geq 3$, $1^2=1$ mod $n$ and $(n-1)^2=1$ mod $n$; $n-1=1$ mod $n$ implies $n=2$ mod $n$, $n-2=kn$ , $kn-(n-2)=(k-1)n+2>2$ if $k>0$ this implies $k=0$ and $n=2$. Contradiction
-
0dont't we have to show that gcd(mm′(modn),1)=1 and 1 <= mm'(modn) < n? – 2017-01-17
-
0you have to show that $gcd(mm',n)=1$. This is i.e to $amm'=kn+1$ i.e $amm'$ =1 mod $n$ and the inverse of $mm'$ mod $n$ is $a$. – 2017-01-17