# Does an Abelian group under multiplication modulo $n$ contain $0$?

I read the following definition in a notebook.

The set $$\mathbb{Z}^{*}_{n}$$ that consists of all integers $$i = 0, 1, ..., n-1$$ for which the $$\operatorname{gcd}(i, n) = 1$$ forms an abelian group under the binary operation multiplicaiton modulo $$n$$. The identity element, of course, is $$1$$.

But I think it should be,

The set $$\mathbb{Z}^{*}_{n}$$ that consists of all integers $$i = 1, ..., n-1$$ for which the $$\operatorname{gcd}(i, n) = 1$$ forms an abelian group under the binary operation multiplicaiton modulo $$n$$. The identity element, of course, is $$1$$.

Or could it be that both the definition are valid?

• Yes, you're right. $0$ wouldn't have an inverse. Jan 31, 2021 at 13:07

Both are valid. Note that $$\operatorname{gcd}(0,n)=n\ne1$$ (for $$n>1$$; but this is not serious restriction). Hence $$0$$ is excluded anyways.
• @DaveIdito Careful; it consists of "all integers $i=0,1,...,n-1$ for which the $\operatorname{gcd}(i,n)=1$". Take $n=4$ and $i=2$ to see that's this does not include all these integers necessarily. Jan 31, 2021 at 13:13
• @DaveIdito there's no difference if you add the given restriction "for which $\gcd(i,n)=1$" (provided $n>1$). For instance, with $n=8$, we do not talak about all the integers $i=1,2,3,4,5,6,7$ either, but only of all of these that are odd. Jan 31, 2021 at 13:13
• No that's not contradictory. On the contrary you have to include $0$ here if you want the statement to be valid for $n=1$. Jan 31, 2021 at 13:14