What are the order $6$ element in $U_7$?
I guess all the members are of order 6 as they are prime to 7.
$U_7$ is cyclic and $\lambda (7)= 6$, where $\lambda$ is the Carmichaeal function. So $a^6 = 1, \forall a \in U_7$.
Is this a correct reason?
And then what happens if $U_n$ is not cyclic, suppose $U_8$.What can you say about order $6$ element in $U_8$. I want some general idea regarding this.
As $\lambda(7)=\phi(7)=6, ord_7a$ will divide $6$ where $(a,7)=1$
$1^1\equiv1\pmod n$ for any natural number $n\implies ord_n1=1$
$-1\not\equiv1\pmod n$ if $n>2$ and $(-1)^2\equiv1\pmod n$
$\implies ord_n(-1)=2 \implies ord_n(n-1)=2$
$2^1\equiv2,2^2\equiv4,2^3=8\equiv1\implies ord_72=3$ and so on
For $U_8, \lambda(8)=2^{3-2}=2\implies ord_8a$ will divide $2$ where $(a,8)=1$
Observe that $(2n+1)^2=8\frac{n(n+1)}2+1\equiv1\pmod 8\implies ord_8(2n+1)$ divides $2$
As $3^1 \not\equiv1\pmod 8 ;ord_83=2$
Similarly, $ ord_85=2$
and $7\equiv-1\pmod 8\implies ord_87=2$
