multiplicative order in field

Question

Let $\alpha$ be primitive element of GF(7). Then order of $\alpha$ is 6, i.e. $o(\alpha)=6$. Now we know that $\alpha^4$ is not equal to 1, and that $o(\alpha^4)$ = $\frac{6}{gcd(6,4)}$. This also can be written as $gcd(6,4)=\frac{6}{o(\alpha^4)}$. This means that $\frac{6}{o(\alpha^4)}$ divides 4.

Now we can write $\alpha^4=\alpha^{Q\frac{6}{o(\alpha^4)}}=(\alpha^6)^{\frac{Q}{o(\alpha^4)}}=1$!!

I am getting this confusion. I might be making an error I am not aware of, it will help if you can look over this and point me in the right direction.

Answer 1

You may be making this a bit more complicated than it actually is. The primitive element $\alpha$ generates a cyclic group of order $6$: $$ \big\langle \alpha \big\rangle = \big\{ 1, \alpha, \alpha^2, \ldots, \alpha^5 \big\} \cong C_6 $$

Now, you seem to be interested in the element $\beta = \alpha^4$ and its order in $GF(7)^\times$. Look at some powers: $$ \begin{align} \beta^2 &= \left( \alpha^4 \right)^2 = \alpha^8 = \alpha^2 \\ \beta^3 &= \beta \beta^2 = \alpha^4 \alpha^2 = \alpha^6 = 1 \end{align} $$

This explicit calculation shows that $$ o(\beta) = o(\alpha^4) = 3. $$

Alternatively, you can arrive at this order by applying the formula: $$ o(\alpha^4) = \frac{6}{\gcd(6, 4)} = \frac{6}{2} = 3. $$

Answer 2

Yes, there is a subtle error. I understand $GF(7)$ to be a field of size $7$, in which case it is isomorphic to $\mathbb{Z}_7$.

From Gallian's book, "Contemporary Abstract Algebra", page 75

$\textbf{Theorem 4.2}$ Let $a$ be an element of order $n$ in a group and lek $k$ be a positive integer. Then $<a^k>=<a^{gcd(n,k)}>$ and $|a^k|=\frac{n}{gcd(n,k)}$.

This does not mean the element $a^k$ equals $a^{gcd(n,k)}$, rather it means they generate the same subgroup. If we apply the theorem and do the substitutions correctly, we get $k=6$ and $n=7$, hence $$|a^4|=\frac{6}{gcd(6,4)}=3$$.

Therefore, because of your argument which is correct, the element $a^4 \in GF(7)$ seems sketchy. The reason is that there is no element of order $6$ in $\mathbb{Z_7}$. The order of any element divides the order of the group, but we have a group of prime order, so they all have order 1 or 7.

The argument is fine, the original assumption is false and hence the contradiction.