# The totient of a factor of a number divides the totient of the number.

Given any number and one of its factors, how can you show that the totient of the factor divides the totient of the original number?

• What do you mean $\phi(mn)=\phi(m)+\phi(n)$? – Hagen von Eitzen Dec 9 '18 at 0:22
• I know that the totient of a number is equal to the sum of the totient of each of its factors so could we not make that statement since mn can be factored into m and n? – Arthur Chong Dec 9 '18 at 0:58

• If $$p\mid n$$ then $$\phi(pn)=p\phi(n)$$.
• If $$p\nmid n$$ then $$\phi(pn)=(p-1)\phi(n)$$.
Expanding on Hagen's answer: Let $$m\mid n$$, we need to show that $$\phi(m)\mid\phi(n)$$. If $$m$$ is prime then we are done, since $$\phi(pm)=p\phi(m)$$. If $$m$$ is not prime, then it can be decomposed into a product of primes, say $$\prod p^\alpha$$. Then
$$\phi(m)=\phi(\prod p^\alpha)=\prod p^{\alpha-1}\prod(p-1).$$
Now, $$\phi(n)$$ can be expressed as this product times the totient function of all the extra prime factors $$n$$ has that is not in $$m$$. Think of this as because we have "taken out" all the prime powers of $$m$$ to get that product, and we can do the same thing for $$n$$ and still have "leftovers". The conclusion follows.