# Is this statement true for every positive integer $n ≥ 2$?

For every positive integer $n ≥ 2$ there exists an element $(a ∈ \mathbb{Z}/n\mathbb{Z})^×$ such that every element of $(\mathbb{Z}/n\mathbb{Z})^×$ is a power of $a$.

Is this true or not and why?

• Try $n=8$, for example. Or $12$. Or $15$. – André Nicolas Oct 26 '15 at 0:54
• No, it's not true. Consider $n = 6$, for which $(\Bbb Z/6\Bbb Z)^{\times} \cong V_4$ – David Wheeler Oct 26 '15 at 0:54
• You are asking whether $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is cyclic. This is not true in general. There is a theorem of Gauss in number theory that characterizes when it is the case. – JohnSmith Oct 26 '15 at 0:55
• I need a little broader help. How is $(Z/nZ)^x$ different than just $(Z/nZ)$ and can you write out as an example why $n = 8$ doesn't work? I just want to see an example so I can understand it better. – Jimm Oct 26 '15 at 0:59

Because you asked for it: the invertible elements (under multiplication mod $n$) of $\Bbb Z/n\Bbb Z$ are precisely the the congruence classes $[a]_n$ for which $\gcd(a,n) = 1$.

It is not hard to see that if $\gcd(a,n) = d > 1$, that we have for $u = \dfrac{n}{d}$, and $t = \dfrac{a}{d}$:

$[u]_n[a]_n = [ua]_n = [u(td)]_n = [(ud)t]_n = [nt]_n = [n]_n[t]_n = _n[t]_n = [0t]_n = _n$.

Thus $[a]_n$ is a zero-divisor, and zero-divisors cannot be invertible.

On the other hand, if $\gcd(a,n) = 1$, then are integers $r,s$ such that:

$ra + sn = 1$, so that, mod $n$:

$[ra + sn]_n = _n$

$[ra]_n + [sn]_n = _n$

$[ra] + _n = _n$

$[r]_n[a]_n = _n$, so that evidently the multiplicative inverse of $[a]_n$ is $[r]_n$.

In particular, let's look at $n = 8$:

Then $(\Bbb Z/8\Bbb Z)^{\times} = \{_8,_8,_8,_8\}$.

Since this group is of order $4$, it is cyclic if and only if it has an element of order $4$. Now the identity ($_8$) has order $1$, while:

$(_8)^2 = [3^2]_8 = _8 = _8$

$(_8)^2 = [5^2]_8 = _8 = _8$

$(_8)^2 = [7^2]_8 = _8 = _8$,

so we have no element of order $4$.

No. The group of units of $\mathbf Z/n\mathbf Z$ is cyclic if and only if $n=2, 4, p^r$ ($p$ being an odd prime) or $2p^r$.