Exercise 2.9.6 (Product of elements of order $3^j$.)

Proof. Since $j>0$, and $a^{3^j}\equiv 1,a^{3^{j-1}}\not\equiv 1(\mathit{mod}p)$, there is some integer $\omega$ (namely $\omega =a^{3^{j-1}}$) such that

(Using the law of quadratic reciprocity (chapter 3), we can prove that the existence of such an $\omega$ is equivalent to $p\equiv 1(\mathit{mod}3)$. We don’t use this result here.)

Then ${\omega }^3-1=(\omega -1)({\omega }^2+\omega +1)$, where $\omega -1\not\equiv 0(\mathit{mod}p)$, thus

Note that ${\omega }^2\not\equiv 1(\mathit{mod}p)$, otherwise $a^{2\cdot 3^{j-1}}\equiv 1(\mathit{mod}p)$, and the order of $a$ is less than $3^j$. Moreover, for every integer $x$,

thus

Here we have chosen $\omega =a^{3^{j-1}}$. Since $b$ has also order $3^j$,

Put $x=3^{j-1}$. Then

Using (1),

Therefore $p\mid (x-\omega )(x-{\omega }^2)$, thus

Now

• If $b^{3^{j-1}}\equiv \omega$, then

  $${(\mathit{ab})}^{3^{j-1}}={\omega }^2,{(a{b}^2)}^{3^{j-1}}={\omega }^3\equiv 1(\mathit{mod}p).$$

  In this case, since ${\omega }^2\not\equiv 1(\mathit{mod}p)$, the order of $\mathit{ab}$ is $3^j$, and the order of $a{b}^2$ is $3^{j^{\prime }}$ for some $j^{\prime }\le j-1$.

• If $b^{3^{j-1}}\equiv {\omega }^2$, then

  $${(\mathit{ab})}^{3^{j-1}}={\omega }^3\equiv 1,{(a{b}^2)}^{3^{j-1}}\equiv {\omega }^5\equiv {\omega }^2(\mathit{mod}p).$$

  Then the order of $a{b}^2$ is $3^j$, and the order of $\mathit{ab}$ is $3^{j^{\prime }}$ for some $j^{\prime }\le j-1$.

To conclude, of the two residue classes $\mathit{ab}$ and $a{b}^2$, one of them has order $3^j$ and the other has order $3^{j^{\prime }}$ for some $j^{\prime }<j$. □