$p$ prime and $m_p$ a proper divisor of $p−1$. Is it $\sigma(m_p)<p$ for every $p$ and $m_p$?
Along the problem I'm facing, I've come to the following lemma (if it is true):
Let $p$ be a prime and $m_p$ a proper divisor of $p-1$. Then the sum of all the divisors of $m_p$, say $\sigma(m_p)$, is less than $p$.
For example, take $p=13$; then, for $m_{13}=6$: $1+2+3+6=12<13$, and likewise for $m_{13}=4,3,2,1$. Or take $p=31$; then for $m_{31}=15$ we have: $1+3+5+15=24<31$, or for $m_{31}=10$: $1+2+5+10=18<31$, and likewise for $m_{31}=5,3,1$.
Is it true for every $p$ and $m_p$?
No. Let $p=37$, $m_{37}=18$. Then $1+2+3+6+9+18 = 39 > 37.$