# Difference between $Z^*_n$ and $Z_n$ [closed]

When studying cyclic groups I stumbled upon the following sentence:

The ∗ in $Z_n^*$ stresses that we are only considering mulitplication and forgetting about addition on this site of stanford.

I don't really understand this definition, as I was learned that $Z_n^*$ just consists out of all elements of the elements of a cyclic group, minus the 0. Where does this the multiplication and addition come in?

## closed as off-topic by fkraiem, tylo, otus, e-sushiMay 29 '17 at 23:32

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• I'm voting to close this question as off-topic because it is about general mathematics. – fkraiem May 29 '17 at 11:21

$\mathbb{Z}_n^*$ doesn't mean $\mathbb{Z}_n - \{0\}$. You must remove all elements that are not invertible mod $n$, which is equivalent to keeping only the elements that are coprimes to $n$.
So, $\mathbb{Z}_n^* = \{x \in \mathbb{Z}_n : \exists x^{-1} \in \mathbb{Z}_n \} = \{x \in \mathbb{Z}_n : gcd(x, n) = 1\}$.
This is necessary since all elements in a group must have inverse elements (regarding the group operation). In $\mathbb{Z}_n$, the group operation is $+$, so, any $x$ has $-x$ as inverse mod $n$. But in $\mathbb{Z}_n^*$, the operation is the multiplication, so, some elements might not have inverses mod $n$.
For instance, take $\mathbb{Z}_9$. The elements $0, 3,$ and $6$ are not invertible mod 9 (we can't multiply them by other element and reduce mod $9$ to obtain $1$). Therefore, $\mathbb{Z}_9^* = \{1, 2, 4, 5, 7\}$.
• Actually, $\mathbb{Z}_n^*$ refers to a group - compared to the ring $\mathbb{Z_n}$ it's not so much about removing elements, it's removing one of the operations and everything that does not fit into the group structure. Everything else follows from the fact that it is a group and by definition every element must have an inverse. – tylo May 29 '17 at 12:35