Asking about some details regarding proof of $5$ having order $2^{n-2}$ in $U(2^n), n \geq3$.

Question

I am asking about the proof that is answer in the following post:

Let $r$ be an integer greater than $2$. Show that $2^r$ divides $\binom{2^{r-2}}{k}2^{2k}$ but it does not divide $\binom{2^{r-1}}{k}2^{2k}$ for $l>2$

There are some statements that are hard to check by myself:

$1)$ The order of $5$ modulo $2^{r+1}$ is a multiple of $2^{r-2}$. (I think we need to check that $2^{r-3}$ is not order)

$2)$ so $5^{2^{r-3}}\equiv 1 + k2^{r-1}\pmod{2^{r}}$, with $k=0$ or $1$.

$3)$ Hence $5^{2^{r-3}}\equiv 1 + 2^{r-1} + k2^r\pmod{2^{r+1}}$, with $k=0$ or $1$.

Everything else is clear and easy. (If no one can give explanations, then I want at least references that can help me to understand these things by myself)

Answer 1

1. The proof is by (strong) induction on $r$. We are assuming the order of $5$ modulo $2^k$ is $2^{k-2}$ for all $k$, $3\leq k\leq r$, and we are trying to prove the order of $5$ modulo $2^{r+1}$ is $2^{r-1}$. That means we are assuming that the order modulo $2^r$ is $2^{r-2}$. So $5^{2^{r-2}}\equiv 1\pmod{2^r}$, but $5^s\not\equiv 1\pmod{2^r}$ for any smaller $s$. So the order modulo $2^{r+1}$ cannot be any smaller than $2^{r-2}$, and whatever the order is modulo $2^{r+1}$, it must be a multiple of the order modulo $2^r$. (For the latter, also: if $5^u\equiv 1\pmod{2^{r+1}}$, then $5^u\equiv 1\pmod{2^r}$; so the order modulo $2^r$ must divide the order modulo $2^{r+1}$). Why would you need to check it is not of order $2^{r-3}$? It cannot be of that order modulo $2^{r+1}$, because it is smaller than the order modulo $2^r$.

2. We are also assuming that $5$ has order $2^{r-3}$ modulo $2^{r-1}$. So $5^{2^{r-3}}\equiv 1\pmod{2^{r-1}}$. That means, by definition, that $2^{r-1}\mid 5^{2^{r-3}}-1$, so there exists $t$ such that $5^{2^{r-3}}=1+t2^{r-1}$. If $t$ is even, then $t2^{r-1}\equiv 0=02^{r-1}\pmod{2^{r}}$. If $t$ is odd, write $t=2u+1$, so $t2^{r-1}= 1\cdot 2^{r-1}+u2^{r}\equiv 1\cdot 2^{r-1}\pmod{2^r}$. Thus, $5^{2^{r-3}} = 1+t2^{r-1}\equiv 1+k2^{r-1}\pmod{2^r}$, with $k=0$ or $k=1$. In the next sentence we show $k=0$ is impossible, so in fact $5^{2^{r-3}}\equiv 1+2^{r-1}\pmod{2^r}$.

3. Repeating the last argument, since $5^{2^{r-3}}\equiv 1+2^{r-1}\pmod{2^r}$, then we will have $5^{2^{r-3}}\equiv 1+2^{r-1} + k2^{r}\pmod{2^{r+1}}$, and we can take $k=0$ or $k=1$.

Answer 2

There's a discussion in $\S6,$ chapter $6$ of Vinogradov's Elements of Number Theory.

The key is that $U(\Bbb Z_{2^r})$ is not cyclic for $r\gt2,$ but rather is $\Bbb Z_2×\Bbb Z_{2^{r-2}}.$

The proof is not by induction and makes use of the binomial theorem.