Suppose $n$ is a Carmichael number.
If, for a contradiction, we supposed that $n=p^k m$ for a prime $p$ and some $k\ge 2$ where $gcd(m,p)=1$. Let $g$ be a generator mod $p^2$.
I am looking to explain why there exists some integer $a$ such that
$$ a=g \; mod\; p^2 $$ $$ and $$ $$a=1 \; mod \; m $$
I know that if $p$ is a prime then there exist $h$ with $1 ≤ h < p$ and $g$ with $1 ≤ g < p^2$ such that $ord_p(h) = p − 1$ and $ord_p^{2} (g) = p(p − 1)$, that is, $h$ is a generator modulo $p$ and g is a generator modulo $p$ however I don't know how to apply this to my question.
Thanks for your help.