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I want to prove that $U(2^n) \simeq \mathbb Z_2 \oplus \mathbb Z_{2^{n-2}}$ for $n \geq3$.

I found this in the site.

The isomorphism is shown by $(-1)^a(5)^b \to (a,b)$ such that $(a,b) \in \mathbb Z_2 \oplus \mathbb Z_{2^{n-2}}$ and $(-1)^a(5)^b \in U(2^n)$ where $a \in \{0,1\}$ and $0 \leq b <2^{n-2}$

I could not understand how $(-1)^a(5)^b$ produce all elements in $U(2^n)$. I know that $|(-1)| \times |5|=2 \times 2^{n-2} = 2^{n-1} = |U(2^n)|$ but i cannot see how being order equivalent causes that result.

So, can you please explain or prove how $(-1)^a(5)^b$ produce all elements in $U(2^n)$, because i cannot see it using the question in link above ?

more explanation

I see that $\langle (-1)\rangle \times\langle 5\rangle $ is generator of $\mathbb Z_2 \oplus \mathbb Z_{2^{n-2}}$, but can see how their combination using $a$ and $b$ $((-1)^a5^b)$ values produce the element in $U(2^n)$. there must be number theory explanation

Thomas Andrews
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  • @DietrichBurde so is it a good reason to downvote a well explained question ? –  Sep 29 '24 at 18:23
  • @DietrichBurde yes they produce the direct group but i cant see how produce $U(2^n)$ –  Sep 29 '24 at 18:24
  • @DietrichBurde yes i understand it but dont understand how the powers of $5$ in both sign produce the elements of $U(2^n)$, because of thati used elementary number theory tags –  Sep 29 '24 at 18:27
  • @DietrichBurde i want to see the last statement in your last comment. what is the relationship between 5 and $U(2^n)$, how it produce its elements –  Sep 29 '24 at 18:28
  • Just because $5$ has order $2^{n-2}$. So $U(2^{n-2})$ has a direct factor of the cyclic group $A$ generated by $5$. But the order is only half the order, so the quotient is isomorphic to $C_2$. Hence $U(2^n)\cong C_2\times A$. This is proved anyway at the duplicate you have linked. – Dietrich Burde Sep 29 '24 at 18:28
  • @DietrichBurde i havent learn quotient groups yet, it is next chapter for me after direct products. so is there any explanation ? –  Sep 29 '24 at 18:30
  • And in the case that $G=A\times B$, the quotient groups is just $G/A\cong B$, or $G/B\cong A$. – Dietrich Burde Sep 29 '24 at 18:37
  • So, what this means for any $n\geq3$ and $a\equiv 1\pmod 4,$ there is a $k$ so that $5^k\equiv a\pmod{2^n}.$ I'd guess trying induction on $n.$ – Thomas Andrews Sep 29 '24 at 18:41
  • Your edit is not correct, it cannot be a generator of $C_2\times C_{2^{n-2}}$, since this group is not cyclic and has no single generator. – Dietrich Burde Sep 29 '24 at 19:33

1 Answers1

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The isomorphism $U(2^n)\cong C_2\times C_{2^{n-2}}$ has been proved already for all $n\ge 3$. The representation as $(-1)^a5^b$ can be seen directly. For this, start with $n=3$. Then $$ U(2^n)=U(8)=\{1,3,5,7\}, $$ and indeed $5^b$ is $5^1=5, 5^2=25=1$ modulo $8$. And $3=-5$, $7=-5^2$. So we are done for $n=3$. Inductively, show that every $k\equiv 1\bmod 4$ can be written as $k=5^b$ modulo $2^n$, and every $k\equiv 3\bmod 4$ as $-5^b$ modulo $2^n$.

Dietrich Burde
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  • can you please explain why we used modulo $4$ –  Oct 07 '24 at 07:56
  • Because of powers of $5$. We have odd numbers, so this is $1$ modulo $2$, like ${1,3,5,7}$, but only $1$ and $5$ are directly powers of $5$. The other two integers $3,7$, which are $3$ modulo $4$, are not powers of $5$, but $-3,-7$ are again. So modulo $4$ is what we need. – Dietrich Burde Oct 07 '24 at 08:15
  • I thought that it is related to every odd number can be written in the form of $4k+1$ or $4k+3$ –  Oct 08 '24 at 15:12
  • Yes, exactly. This is what I said about odd numbers $1,3,5,7$ and so on. Only the ones congruent to $1$ modulo $4$ are powers of $5$. So we have the natural distinction of odd numbers of the form $4k+1$, or $4k+3$, but this is only useful, because it is linked to powers of $5$, our generator for the cyclic group $C_{2^{n-2}}$. – Dietrich Burde Oct 08 '24 at 15:51