TITLE: A proof involving the Euler phi function
QUESTION [11 upvotes]: Problem:
Let $\varphi$ be the Euler phi function, where for any $n \in \mathbb{Z^+}$, $\varphi(n)$ is the number of positive integers less than $n$ that are relatively prime with $n$.
Show that if $d \mid n$, then $\varphi(d) \mid \varphi(n)$, where $d$ and $n$ are positive integers.

My attempt:
Because $d \mid n$, there exist some $a \in \mathbb{Z}$ such that $da = n$.
If $d = 1$, then $\varphi(d) = 1$, so $\varphi(d) \mid \varphi(n)$ for any $n \in \mathbb{Z^+}$ and we are done.
So instead, suppose that $d \neq 1$. Now I'm trying to show that $\{\Delta \in \mathbb{Z^+}\mid \Delta < d, \ \gcd(\Delta, d) = 1\}\subseteq \{\Delta \in \mathbb{Z^+}\mid \Delta < d, \ \gcd(\Delta, n) = 1\}$ by showing that whenever an integer $\Delta < d$ is relatively prime with $d$ (so that it is "inside" $\varphi(d)$), then it is also relatively prime with $n$.
However, I realised that this is not possible to show. If $n = \Delta \cdot d$, then the premise $d \mid n$ is true, but if $\gcd(\Delta, d) = 1$,  then $\gcd(\Delta, n) = \Delta$, so the subset relation from a few lines above cannot be true in this situation. So I think this can't be the right approach. If someone can please give me a pointer, I will be very grateful.

REPLY [5 votes]: Let,  
$$d={p_{1}}^{k_{1}}{p_{2}}^{k_{2}} \cdot \cdot\cdot {p_{j}}^{k_{j}}$$   
$\text{&}$  
$$n={p_{1}}^{l_{1}}{p_{2}}^{l_{2}} \cdot \cdot\cdot {p_{j}}^{l_{j}}{p_{j+1}}^{l_{j+1}} \cdot \cdot \cdot {p_{r}}^{l_{r}} $$   
Where $\{p_{i}\}_{i=1}^r$ are prime numbers and $(k,j,r) \in \mathbb W$
Since $ \ d|n \implies l_{i} \geq k_{i} \ \ \forall \ \ i \in \{1,2,3,\cdot\cdot\cdot,j\} $  
From the definition of $\phi$ function, we have,   
$$\phi(d)={p_{1}}^{(k_{1}-1)}{p_{2}}^{(k_{2}-1)} \cdot \cdot\cdot {p_{j}}^{(k_{j}-1)} \prod_{i=1}^{j} (p_{i}-1) $$  
$\text{&}$  
$$\phi(n)={p_{1}}^{(l_{1}-1)}{p_{2}}^{(l_{2}-1)} \cdot \cdot\cdot {p_{r}}^{(l_{r}-1)} \prod_{i=1}^{r} (p_{i}-1) $$     
Since $l_{i} \geq k_{i} \implies l_{i} -1 \geq k_{i} -1  \ \ \forall \ \ i \in \{1,2,3,\cdot\cdot\cdot,j\} $. Also, $r \geq j$  
$\implies \phi(d)|\phi(n)$  
Q.E.D.