# How can I show that $\phi(m) \mid \phi(n)$? [duplicate]

I want to prove that:

$$\text{ if } m,n \geq 1 \text{ and } m \mid n,\text{ then } \phi(m) \mid \phi(n).$$

How can I show this?

I thought the following:

$$m \mid n \Rightarrow \exists k \in \mathbb{Z} \text{ such that } n=km$$

$$\text{We write } m \text{ like that : }m=p_1^{a_1} \cdots p_k^{a_k}, \text{ where } p_i \text{ are primes and } a_i>0$$

Then,we have $n=k p_1^{a_1} \cdots p_k^{a_k}$

We could show that $\phi(m) \mid \phi(n)$,using the fact that as $\phi$ is multiplicative and if $(m,n)=1$ ,then $\phi(mn)=\phi(m) \phi(n)$

But..it is possible that $k \mid p_1^{a_1 \cdots p_k^{a_k}}$.

So,do I have to show maybe that $\gcd(kp_1^{a_1} \cdots p_{k-1}^{a_{k-1}},p_k^{a_k})=1$ ?

## marked as duplicate by Marc van Leeuwen, Namaste, user91500, Gabriel Romon, vonbrandJun 20 '14 at 17:42

• Your notation has a bad aspect: you use $k$ both for the complementary factor of $m$ in $n$ and for the number of distinct prime factors of $m$. Just look at the expression $kp_1^{a_1}\cdots p_k^{a_k}$ that you wrote. – KCd Jun 20 '14 at 13:45

You could use the fact that if $n = p_1^{a_1}\ldots p_k^{a_k}$ then $\phi(n) = (p_1 - 1)p_1^{a_1 -1}\ldots (p_k - 1)p_k^{a_k -1}$

• So,can I say it like that? $$\text{Let } n=p_1^{a_1} \cdots p_k^{a_k}$$ Then $\displaystyle{\phi(n)=p_1^{a_1-1}(p_1-1) \cdots p_k^{a_k-1}(p_k-1)}$ $$m \mid n \Rightarrow m \mid p_1^{a_1} \cdots p_k^{a_k}$$ $$m \geq 1, \text{ so it has a prime divisor, let } q$$ $$q \mid m, m \mid p_1^{a_1} \cdots p_k^{a_k} \Rightarrow q \mid p_1^{a_1} \cdots p_k^{a_k} \Rightarrow q \mid p_i, \text{ for an } i \in \{ 1, \dots,k\}$$ Let $q \mid p_1 \Rightarrow q=p_1$ So, $m=l \cdot p_1 \text{ and then: } \phi(m)=\phi(p_1^{b_1} \cdot q_2^{b_2} \cdots q_j^{b_j})=p_1^{b_1-1}(p_1-1)q_2^{b_2} \cdots q_j^{b_j}$ – evinda Jun 20 '14 at 13:29
• You have that $m|n$. So $m = p_1^{b_1}\ldots p_j^{b_j}$ where $b_i \le a_i$ for every $i$. You can calculate $\phi(m)$ exactly as we did for $\phi(n)$, and it will be clear that $\phi(m)|\phi(n)$. The key is that since $m|n$, all the prime divisors of $m$ divide $n$. – Mathmo123 Jun 20 '14 at 13:36
• So,it is like that: $$\phi(m)=p_1^{b_1-1} \cdots p_k^{b_k-1}(p_1-1) \cdots (p_k-1)$$ And,since $\displaystyle{ \frac{p_1^{a_1-1} \cdots p_k^{a_k-1}(p_1-1) \cdots (p_k-1)}{p_1^{b_1-1} \cdots p_k^{b_k-1}(p_1-1) \cdots (p_k-1)}=p_1^{a_1-b_1} \cdots p_k^{a_k-b_k} \in \mathbb{Z} (a_i \geq b_i, \forall i \in \{1, \dots,k \}),\text{ we conclude that } \phi(m) \mid \phi(n). }$ Right?? – evinda Jun 20 '14 at 13:48
• That's correct. – Mathmo123 Jun 20 '14 at 13:54
• For a better idea of what's going on though, I'd suggest looking at @KCd's answer too. – Mathmo123 Jun 20 '14 at 13:57

Here is a more conceptual viewpoint, which is not how you are thinking about the problem, but I think it provides a nice explanation for the result without grinding out formulas for the $\varphi$-function. You need to know some abstract algebra to makes sense of what is below, or this may be some motivation to learn it.

Since $m|n$, there is a natural reduction map ${\mathbf Z}/n{\mathbf Z} \rightarrow {\mathbf Z}/m{\mathbf Z}$, given by $a \bmod n \mapsto a \bmod m$, which is a ring homomorphism. It is obviously surjective. This ring homomorphism induces a homomorphism on the unit groups $(\mathbf Z/n\mathbf Z)^\times \rightarrow (\mathbf Z/m\mathbf Z)^\times$. Prove, less obviously, that this map between unit groups is surjective. (Hint: think about the Chinese remainder theorem, if you know what that is.) Then you are done, since if there is a surjective homomorphism between finite groups $H \rightarrow G$ then necessarily $|G|$ divides $|H|$.

The basic point is that if you are trying to show one positive integer $a$ divides another positive integer $b$, you might try to explain it with group theory: find a finite group $G$ with order $a$ and a finite group $H$ with order $b$, and either construct an injective group homomorphism $G \rightarrow H$ or a surjective group homomorphism $H \rightarrow G$. Either way it would follow that $a$ divides $b$ (by Lagrange's theorem for the injective case, and by quotient groups and the first homomorphism theorem in the surjective case). In your problem $a = \varphi(m)$ and $b = \varphi(n)$, which are the orders of the unit groups mod $m$ and mod $n$.

• That the map of the unit groups is surjective is actually just about as hard as this theorem. – Thomas Andrews Jun 20 '14 at 13:54
• Oh, I would say it is harder than the question being asked! But it explains the result in a nice way, and for someone who has had a bit of algebra and number theory I think it is an instructive task to prove the result by such a method without using formulas for the $\varphi$-function. Note also that I did acknowledge at the start that this approach is not how the OP has been trying to tackle the problem. It is probably more suitable at the moment for students with more background than the OP. – KCd Jun 20 '14 at 13:58
• This is essentially the same as this answer to the duplicate question, and the problem that I mentioned there in a comment is the same as the one signalled by @ThomasAndrews. Talk about déjà vu! Don't take this as as critique: I just gave an answer myself that duplicates my answer there (but with errors); yet it was easier to answer than to find the duplicate. – Marc van Leeuwen Jun 20 '14 at 15:15
• @MarcvanLeeuwen: A difference is that in my answer I acknowledge that one really does have to show that the map on unit groups is surjective. – KCd Jun 20 '14 at 16:40

If the highest power of prime $p_i$ in $m,n$ are $M_i,N_i$ respectively, $0\le M_i\le N_i$ for all $i$

Now for each $M_i>0,$ $$\displaystyle\phi(m)=\prod\phi(p_i^{M_i})=\prod p^{M_i-1}(p_i-1)$$

and similarly for each $N_i>0,$ $$\displaystyle\phi(n)=\prod\phi(p_i^{N_i})=\prod p^{N_i-1}(p_i-1)$$

$\displaystyle p^{M_i-1}$ clearly divides $p^{N_i-1}$ for all $i$ as $M_i\le N_i$ for all $i$

So, the proposition follows

• "is clearly divisible by" should read "clearly divides". – TonyK Jun 20 '14 at 13:18
• @TonyK, Thanks for your observation. Rectified – lab bhattacharjee Jun 20 '14 at 13:19
• @labbhattacharjee Could you explain it further to me? I haven't understood it.. Also,do you maybe know how I could continue,doing it with the following way? – evinda Jun 20 '14 at 13:34
• So,can I say it like that? $$\text{Let } n=p_1^{a_1} \cdots p_k^{a_k}$$ Then $\displaystyle{\phi(n)=p_1^{a_1-1}(p_1-1) \cdots p_k^{a_k-1}(p_k-1)}$ $$m \mid n \Rightarrow m \mid p_1^{a_1} \cdots p_k^{a_k}$$ $$m \geq 1, \text{ so it has a prime divisor, let } q$$ $$q \mid m, m \mid p_1^{a_1} \cdots p_k^{a_k} \Rightarrow q \mid p_1^{a_1} \cdots p_k^{a_k} \Rightarrow q \mid p_i, \text{ for an } i \in \{ 1, \dots,k\}$$ Let $q \mid p_1 \Rightarrow q=p_1$ So, $m=l \cdot p_1 \text{ and then: } \phi(m)=\phi(p_1^{b_1} \cdot q_2^{b_2} \cdots q_j^{b_j})=p_1^{b_1-1}(p_1-1)q_2^{b_2} \cdots q_j^{b_j}$ – evinda Jun 20 '14 at 13:35
• @evinda, Please find the edited version – lab bhattacharjee Jun 20 '14 at 13:37

We have $$\phi(n)=n\prod_{p\mid n}\left(1-\frac1p\right)$$ $$\phi(m)=m\prod_{p\mid m}\left(1-\frac1p\right)$$

Now $m\mid n$ implies that the set of primes $p\mid m$ is a subset of the primes $p\mid n$. Hence $\phi(n)/\phi(m)$ is an integer because the factors corresponding to the common primes cancel and the denominators of the other factors are cancelled by $n$.

• Shouldn't the first product subscript be $p \mid n$? This is a rather unconvincing argument, as $\phi(n)$ includes more fractional factors than $\phi(m)$ in the above representation. – Erick Wong Jun 20 '14 at 14:23
• @ErickWong, thanks. See my edited answer. – lhf Jun 20 '14 at 14:35

Since $\phi$ is a multiplicative function, this question reduced to the case where $n,m$ are powers of the same prime. Since $\phi(p^k)=(p-1)^{\min(k,1)}p^{\max(k,1)-1}$, where both $k\mapsto\min(k,1)$ and $k\mapsto\max(k,1)-1$ are increasing functions $\Bbb N\to\Bbb N$, the result holds in the mentioned case.