Question:
Find a subgroup of order 6 in $U\left ( 700 \right )$.
Observe that $700=7.25.4$
Indeed, $gcd\left ( 7,25,4 \right )$ is $1$.
Hence, $U\left ( 7.100 \right )\cong U\left ( 7 \right ) \bigoplus U\left ( 25\right ) \bigoplus U\left ( 4 \right )$.
Noting that the order of $U\left ( 7 \right )$ is $6$, the order of $U\left ( 25\right )$ is $20$ and the order of $U\left ( 4 \right )$ is $3$, we note that the order of the external direct product is $360$.
By the fundamental theorem of cyclic group: the order of the subgroups of a group is a divisor $k$ of $n$.
Thus, $k = \left \{ 1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180,360 \right \}$.
A unique subgroup of order 6 of $U\left ( 700 \right )$ is $\left \langle a^{60} \right \rangle, \exists a \in U\left ( 7 \right ) \bigoplus U\left ( 25\right ) \bigoplus U\left ( 4 \right ).$
I think the general stratagem of my attempt is in the right correction but it does feel as though my attempt isn't complete. Can I determine $a$?
Hints are more appreciated.
$U(n)$ is the group of units of the integers modulo $n$, that is it consists of elements $[x]$ with $\gcd(x,n)=1$, $[x]=[y]$ when $x\equiv y\pmod n$ and $[x][y]=[xy]$, right?
The $U(7)$ in your decomposition consists of the elements $[x]$ of $U(700)$ with $x\equiv1\pmod{100}$, that is it consists of $[1]$, $[101]$, $[201]$, $[401]$, $[501]$ and $[601]$. Have any of these order $6$ in $U(700)$?