Multiplication of three primitive roots

Question

I have noticed that if I multiply three primitive roots of the same modulo it is still a primitive root in that modulo.
But I cant manage to prove it or this isn't true?
Let $x,y,z$ be primitive roots of modulo $n$, is $xyz$ mod $n$ necessarily a primitive root?

Answer 1

Modulo 19, have $2, 3, 14$ primitive roots, but their product is $8 \pmod{19}$ which is not a primitive root because $8^6 \equiv 1 \pmod{19}$.

In fact $n=19$ is the first time this happens with distinct primitive roots, and $10, 13, 15$ is the only other example with $n=19$.

Answer 2

Consider $n=7$, suppose $x$ is a primitive root. Let $y=z=x$. Then $xyz=x^3$ is not a primitive root since $gcd(3,n-1)\neq 1$.

Answer 3

Generalization of Patrick's post:

Choose any prime of the form $p=6(2m+1)+1$

As ord$_ma=d,$ord$_m(a^k)=\dfrac d{(d,k)}$ (Proof @Page$\#95$ of George E. Andrews' Number Theory)

So, if $a$ is a primitive root $\pmod p,$ so will be $a^{6r+1}$

If $r\ge2,6r+1\ge13$ and we need $13\le6r+1\le\phi(p)$

Now for $b=a\cdot a^{6+1}\cdot a^{6\cdot2+1}=a^{21},$ ord$_p(b)=\dfrac{p-1}{(p-1,21)}=\dfrac{p-1}{(12m+6,21)}=\dfrac{p-1}{3(4m+2,7)}<p-1$

Answer 4

Fix a primitive root, $g$. Then your three chosen ones can be written as $g^n,g^m,g^k$ for suitable exponents. The product is then $g^{n+m+k}$. That product is again a primitive root iff $\gcd (n+m+k,p-1)=1$. If, for example, $p\equiv 1 \pmod 3$ and we take the three roots all equal to $g$ then the product $g^3$ is never a primitive root.

Answer 5

Fix a primitive root, $g$. Then your three chosen ones can be written as $g^n,g^m,g^k$ for suitable exponents. The product is then $g^{n+m+k}$. That product is again a primitive root iff $\gcd (n+m+k,p-1)=1$. If, for example, $p\equiv 1 \pmod 3$ and we take the three roots equal to $g$ then the product $g^3$ is never a primitive root.