What is the size of the multiplicative group $(\mathbb{Z}/p^n\Bbb{Z})^ \times$ if p is a prime number?
At first, I thought that the size would be $p^n-1$, but now I think perhaps not because $(\mathbb{Z}/p^n\Bbb{Z})$ might have more than one non-invertible element.
On
The non-units is $\mathbf Z/p^n\mathbf Z\;$ are the congruence classes of the integers which are divisible by $p$. Representatives of each of these classes are obtained as $\bigl\{kp\mid 0\le k \le p^{n-1}-1\bigr\}\;$ (or $1\le k\le p^{n-1}$), which are $p^{n-1}$.
Hence the number of units is $\;p^n-p^{n-1}=p^{n-1}(p-1)$.
It's a well known result (since it can be shown that $|(\mathbb Z/n\mathbb Z)^\times|=\phi(n))$ that $$|(\mathbb Z/n\mathbb Z)^\times|=n\prod_{p|n}\bigg(1-\frac1p\bigg)$$Where $p$ are primes that divide $n$.
Note that for $n = p^k$, only $p$ is the prime that divides $n$. As such, $$|(\mathbb Z/p^k\mathbb Z)^\times|=p^k\bigg(1-\frac1p\bigg)=p^k-p^{k-1}$$