Property of modulo congruation

Question

If I have: $$a^b \equiv 1 \mod xy$$ where $x,y$ are primes, is then true that: $$ a^b \equiv 1 \mod x$$ $$ a^b \equiv 1 \mod y$$

I don't sure if this is true, because I don't know how can I prove it

Answer 1

$x = 1 \pmod {y(z+1)} \implies x = 1 \pmod {z+1}, \ = 1 \pmod y$. Similarly, $ x = 1 \pmod {y(z+k)} \implies x = 1 \pmod{y}, \pmod{z+k}$. Simlarly for any polynomial in $\Bbb{Z}[x_1, \dots, x_k], \ \ p(\bar{x}) = \ p_1(\bar{x}) \cdots p_n(\bar{x}), $ we have that if $z = z' \pmod {p(\bar{y})}$, then $z = z' \pmod{p_i(\bar{y})}, \ \forall i=1\dots n$.