Divisibility of Euler's totient function: if $m|n$ then $\phi(m)|\phi(n)$

Question

This problem is from Gerstein's Introduction to Mathematical Structures and Proofs: show that if $m|n$ then $\phi(m)|\phi(n)$. One way to approach this is to use the prime factorizations of m and n:

$m=\Pi p_i^{e_i}$

$n=\Pi p_j^{f_j}$

and argue that the $p_j$ include all the $p_i$ with $f_j\ge e_i$

Then

$\phi(m)=\Pi p_i^{e_i-1}\Pi(p_i-1)$

and

$\phi(n)=\Pi p_j^{f_j-1}\Pi(p_j-1)$

You can kind of see that all the factors of $\phi(m)$ are also present in $\phi(n)$, but is there a more direct proof?

I’d call that a very direct proof. Unique factorization tells you that $n$ has all of the prime factors of $m$ with at least their multiplicities in $m$, and Euler’s formula then gives you the result directly. — Brian M. Scott, Aug 27 '20 at 15:59
Same question here. Do any of those answers count as a "direct proof"? See also here. — Bill Dubuque, Aug 27 '20 at 16:02
You folks are right, I think. My proof is easier to understand than the one cited by @Gone — Anna Naden, Aug 27 '20 at 16:17
@Anna If you rigorously write out the details of "kind of see that" then you essentially get the linked proof(s). — Bill Dubuque, Aug 27 '20 at 16:42

score 2 · Answer 1 · answered Aug 27 '20 at 16:00

The factorizations of $\phi(m)$ and $\phi(n)$ provide a direct proof.

Since if $e_i\leq f_i$, then $e_i-1\leq f_i-1$, and if $p_i$ divides $m$, then $p_i$ divides $n$. This shows that

$\prod_i p^{e_i-1}\prod_i(p_i-1)$ divides $\prod_j p^{e_j-1}\prod_j(p_j-1)$.

Divisibility of Euler's totient function: if $m|n$ then $\phi(m)|\phi(n)$

1 Answers1