# $\sum_{d|n} \varphi(d)=n$

I want to solve $\sum_{d|n} \varphi(d)=n$ using Group theory.

Here, $\varphi(d)$ is Euler's totient function.

I think I should use $\Bbb Z_n$ and fundamental theorem of cyclic group.

Then I use $\varphi(d)$ as the number of generators of $\Bbb Z_d$

But I can't link this idea with the sum of all $d|n$

Help me!

• The order of a subgroup of the cyclic group is a divisor of $n$ and there is exactly one such group for each divisor. – Winther Apr 3 '16 at 14:56
• Is math.stackexchange.com/questions/948620/… not doing it for you? – Pedro Tamaroff Apr 3 '16 at 14:56
• @PedroTamaroff Thank you!! I understand it – Pearl Apr 3 '16 at 15:18

Hint: use the fact that, if $n=\prod p_i^{\alpha_i}, \mathbb {Z_n}\simeq \prod \mathbb {Z_{p_i^{\alpha_i}}}$.