If a prime $p \mid n$, then why does $p - 1 \mid \phi(n)$?

Question

I saw in the top answer for this question Show that there is no integer n with $\phi(n)$ = 14 that the following assumption was made:

If a prime $p \mid n$, then $p - 1 \mid \phi(n)$, with $\phi(n)$ referring to Euler's totient function applied to $n$.

However, I am confused as to why this assumption is true.

The Euler phi function is multiplicative. Any decent introduction to the subject will show how a prime factorization of $n$ gives an expression for $\varphi(n)$. — hardmath, Nov 14 '21 at 06:00

score 1 · Accepted Answer · answered Nov 14 '21 at 06:02

1

If $p \mid n$, then $n = p^v\cdot m$ (where $p\nmid m)$. Then $$\phi(n) = \phi(p^v \cdot m) = \phi(p^v)\cdot \phi(m) = (p-1)p^{v-1}\phi(m)$$ Therefore, $p-1 \mid \phi(n)$.

answered Nov 14 '21 at 06:02

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If a prime $p \mid n$, then why does $p - 1 \mid \phi(n)$?

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