I have been stuck on this question for a long time and don't really understand how the Chinese remainder the is related to the order of a unit.
Use the Chinese Remainder Theorem to find an element of order 12 in G = (Z/105Z)^×. Are there any elements of larger order in G?
Here G is the multiplicative ring mod 105.
On
By the Chinese Remainder theorem, we have that $$\bigl(\mathbf Z/105\mathbf Z\bigr)^{\!\times}\simeq\bigl(\mathbf Z/3\mathbf Z\bigr)^{\!\times}\times\bigl(\mathbf Z/5\mathbf Z\bigr)^{\!\times}\times\bigl(\mathbf Z/7\mathbf Z\bigr)^{\!\times}$$ As the order of an element $(x,y,z)$ in the product is the l.c.m. of the orders of $a,b,c$. By Lagrange's theorem, the order of $a$ is a divisor of $2$, the order of $b$ a divisor of $4$ ands the order of $c$ a divisor of $6$, so it is enough to take $a=1$, $b$ of order $4$, e.g. $b=2$, and $c$ of order $6$, say $c=3$.
We first solve $\; x\equiv 1\mod 3,\quad x\equiv 2\mod 5$.
A Bézout relation between $3$ and $5$ is: $\;2\cdot 3 -5=1$, hence the solutions are $$x\equiv 2\cdot 6-1\cdot 5=7\mod 15.$$
Now let's solve: $\; x\equiv 7\mod 15,\quad x\equiv 3\mod 7$.
A Bézout relation between $15$ and $7$ is: $\;15-2\cdot 7=1$, hence the solutions are: $$x\equiv 3\cdot 15-7\cdot14=-53\equiv \color{red}{52}\mod 105.$$
Hint: Note that $3$ has order $6$ modulo $7$, and $2$ has order $4$ modulo $5$. Solve the simultaneous system of congruences $x\equiv 3\pmod{7}$, $x\equiv 2\pmod{5}$.