Today in my group theory class, my teacher was proving the following statement:
If $n \in \mathbb N$ and if $n = \prod_{k = 1}^t p_k ^{\alpha_k}$ is it's prime factorization, then: $$\varphi(n) = n\prod_{p\mid n } \left(1 - \frac {1}{p} \right)$$ where $p$ is prime.
Here $\varphi$ is the Euler's totient function. The proof he presented is the following:
Let $\mathbb Z^*_n$ denote the multiplicative group of integers mod $n$.
If $m,n \in \mathbb N$ such that $\gcd(m,n) = 1$, then:
$$\mathbb Z^*_n \times \mathbb Z^*_m \simeq \mathbb Z^*_{nm}$$
This means that $| \mathbb Z^*_{nm}| = | \mathbb Z^*_{n}|| \mathbb Z^*_{m}|$, thus:
if $m,n \in \mathbb N$ such that $\gcd(m,n) = 1$, then: $$\varphi(nm) =\varphi(n)\varphi(m)$$
So, if $n = \prod_{k = 1}^t p_k ^{\alpha_k}$, then: $$\varphi(n) = \prod_{k = 1}^t \varphi (p_k ^{\alpha_k})$$
We have that if $p$ is prime then:
$$\varphi(p^k) = p^{k - 1}(p-1)$$
So, putting all together we have that:
$$\varphi(n) = \prod_{k = 1}^t \varphi (p_k ^{\alpha_k}) = \prod_{k = 1}^t p_k^{\alpha_k - 1}(p_k-1)$$
And then the teacher claimed that $$\prod_{k = 1}^t p_k^{\alpha_k - 1}(p_k-1) = n \prod_{p\mid n} \left(1 - \frac {1}{p} \right),$$
where $p$ is prime. This is the only step of the proof that my teacher just wrote on the board without a proper explanation and that's why I'm having some struggle understanding it. Why is this true?
It's a product, so you can take out a factor of $n$ as a product of primes. Then, by definition, the $p_k$ are the $p|n$.
$$\begin{align} \prod_{k = 1}^t p_k^{\alpha_k - 1}(p_k-1)& = \prod_{k = 1}^t p_k^{\alpha_k}\left(1 - \frac {1}{p_k} \right)\\ & = \left(\prod_{k = 1}^t p_k^{\alpha_k}\right)\prod_{k = 1}^t\left(1 - \frac {1}{p_k} \right) \\ &= n \prod_{p|n} \left(1 - \frac {1}{p} \right) \end{align}$$