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so I am trying to find out how to prove that 3 is a primitive root of $7^k$ for all $k \ge 1$. I am trying to prove this via induction. Thanks.

terminix00
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  • Can you prove it for k=1? – Jonas Gomes Apr 15 '14 at 02:00
  • Related : http://math.stackexchange.com/questions/594782/2-is-a-primitive-root-mod-3h-for-any-positive-integer-h and http://math.stackexchange.com/questions/332760/prove-that-a-primitive-root-of-p2-is-also-a-primitive-root-of-pn-for-n1?lq=1 – lab bhattacharjee Oct 14 '14 at 08:11

2 Answers2

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The solution depends on how much theory we have available. The standard theorem here is that if $a$ is a primitive root of $p^2$, where $p$ is prime, then $a$ is a primitive root of $p^k$ for any $k\ge 2$.

So we need only verify that $3$ is a primitive root of $7^2$. This is in principle a computation, but we can speed it up. It is easy to verify directly that $3$ is a primitive root of $7$. For $7^2$, we need to show that $3$ has order $\varphi(7^2)=42$ modulo $7^2$. The possible orders are multiples of $6$ that divide $42$. so we need only show that $3$ does not have order $6$ modulo $7^2$. Calculate. We have $3^6=729\equiv 43\pmod{49}$.

André Nicolas
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  • I see. I guess this leads me to my other question. How come we only check $3^6$ and not $3^7$ or any other divisors of 42? – terminix00 Apr 15 '14 at 02:23
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    The order of $3$ modulo $7$ is $6$. If $a^n \equiv 1\pmod{7^2}$, then $a^n\equiv 1\pmod{7}$, and therefore $6$ divides $n$. If it did not, let $n=6q+r$ where $1\le r\le 5$. Then $1\equiv 3^n\equiv(3^6)^q3^r\equiv 3^r\pmod{7}$, contradicting the fact the order of $3$ modulo $7$ is $6$. – André Nicolas Apr 15 '14 at 02:51
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Just for semi-completeness, a proof of Andre's claim.

First, use that the multiplicative group $\mathbb Z_{p^k}^\times$ is cyclic.

So the number of generators is $a_k=\phi(p-1)(p-1)p^{k-2}$ for $k\geq 2$.

So that means that if $a_{k+1}=pa_k$. But the only generators modulo $p^{k+1}$ are generators modulo $p^k$, and there are only $pa_k$ numbers, modulo $p^{k+1}$ which are congruent to generators modulo $p^{k}$. So that means all such elements must be generators.

So this follows if we know that $\mathbb Z_{p^k}^\times$ is cyclic.

Thomas Andrews
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