# Let the divisors of $p-1$ be $d_1,d_2,\cdots$Let $g \mod p$ be a primitive root ,then for each $d_i$ there is an element with period $d_i$

I am trying,without success,to prove this statement:

Let the divisors of $p-1$ be $d_1,d_2,\cdots$ Prove that if we have a primitive root $g \mod p$ ,then for each $d_i$ there is an element with period $d_i$

I am trying,without success,to prove this statement:

Let the divisors of $p-1$ be $d_1,d_2,\cdots$ Prove that if we have a primitive root $g \mod p$ ,then for each $d_i$ there is an element with period $d_i$

My thinking:

It's given that $g$ is a primitive root ,now let $g=a^t$ where $a$ is not a multiple of $p$ and $t$ is some real number ,so we have that $a^{t},a^{2t},a^{3t},\cdots \equiv 1 \mod p$

Now I know that from Fermat we have $a^{p-1} \equiv 1 \mod p$, so $p-1=t\cdot k$ for some integer $k$.

Since we have that each of $d_i$ is a divisor of $p-1$ we have that $p-1=d_i \cdot q$ where $q$ is an integer,then $d_i \cdot q =t\cdot k$ so I have that $$a^{d_i}=a^{tk/q} =(a^{tk})^{1/q}$$

Since we have $a^{tk}\equiv 1 \mod p$ we have also that $\left(a^{tk}\right)^{1/q} \equiv 1^{1/q} \equiv 1 \mod p$

Now I don't know what should I do now ,I am terribly confused (I haven't someone to ask for advice)

• What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something. Jan 15, 2016 at 13:24
• Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 \mod p$ Jan 15, 2016 at 13:34
• In that case, $g=a^t$ never holds for $p>2$, because $a^t\equiv 1$ and $g\not\equiv 1$. Jan 15, 2016 at 13:37
• Can't I have $g=a^{t}$ for some real $t$ ? Jan 15, 2016 at 13:45
• It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $a\equiv b\mod p$, then $a^t=b^t\mod p$. Instead, I would suggest you looking at the integer powers of $g$ itself. Jan 15, 2016 at 13:56

It's a much more general fact. If $$g$$ is an element of period $$n$$ in a group, and $$m$$ divides $$n$$, then the period of $$g^{n/m}$$ is $$m$$.
Even more generally, when $$k$$ is arbitrary, then the period of $$g^{k}$$ is $$\frac{n}{\gcd(n, k)}.$$
For $$1 \le k < m$$, we have $$(n/m) \cdot k < n$$, so $$(g^{n/m})^{k} = g^{(n/m)\cdot k} \ne 1$$. However $$(g^{n/m})^{m} = g^{(n/m) \cdot m} = g^{n} = 1$$, so the period of $$g^{n/m}$$ is exactly $$m$$.