Proof that in the unit group $U_{2^\alpha}$ there is an element of order $2^{\alpha-2}$.

Question

Let's consider the unit group $U_{2^{\alpha}}$, that is not cyclic for $\alpha>1$. We are sure that there isn't any element with order $2^{\alpha-1}$ (or the group would be cyclic). Can we demonstrate that is there always an element of order $2^{\alpha-2}$?

Answer 1

Let's try induction. We need to show that, when we write $5^{2^{\alpha-2}}-1=k2^\alpha$, we have $k$ odd. Why is this necessary? Well, it establishes that the $2^{\alpha-2}$ power of $5$ is congruent to $1$ modulo $2^{\alpha}$, but that it is not congruent to $1$ modulo $2^{\alpha+1}$. Looking at specific values of $\alpha$, we get that $5^2$ is congruent to $1$ modulo $8$, but since it's an odd multiple of $8$, plus $1$, it's not congruent to $1$ modulo $16$. Similarly, $5^4$ is congruent to $1$ modulo $16$, but not modulo $32$, because it equals $39\times16+1$. Thus, the order of $5$ as an element of the unit group keeps increasing for successive powers of $2$.

For $\alpha=3$, our claim holds, because $25-1=3\times 8$. Now, we write $$5^{2^{\alpha-2}}=k2^\alpha+1,$$ and we square both sides: $$5^{2^{\alpha-1}}=k^22^{2\alpha}+k2^{\alpha+1}+1,$$ then we move the $1$ back over and factor the right: $$5^{2^{\alpha-1}}-1=(k^22^{\alpha-1}+k)2^{\alpha+1}.$$

When $k$ is odd, then so is $(k^22^{\alpha-1}+k)$, so the induction step is complete.

Does that work?

Answer 2

Yes, $5$ has order $2^{\alpha-2}$ mod $2^{\alpha}$. Just prove by induction that $5^{2^{\alpha-3}} \equiv 1 + 2^{\alpha-1}\bmod 2^{\alpha}$.