How to prove that $U_{2^n}$ is isomorphic as group to $\mathbb Z_2 \times \mathbb Z_{2^{n-2}}$ for $n \ge 3$ ?
Asked
Active
Viewed 3,445 times
7
-
1What is it you denote by $U_{2^n}$ ? The unitary group $\lbrace z\in {\mathbb C} | z^{2^n}=1\rbrace$ ? – Ewan Delanoy Jun 01 '15 at 14:27
-
@Ewan Delanoy : No , it is the group of units of $\mathbb Z_{2^n}$ – Jun 01 '15 at 14:29
-
1Does this answer your question? The structure of the group $(\mathbb{Z}/2^n\mathbb{Z})^*$ – Bill Dubuque May 28 '23 at 19:23
2 Answers
6
Hint : Show that for $n \geq 3$, $5$ has order $2^{n-2}$, arguing as follows: by induction, you can prove that $5^{2^{n-3}} \equiv 1 + 2^{n-1}$ mod $2^{n}$. Now you know that $-1 \equiv 2^{n}-1$ has order 2. Using this, you can prove that every element can be written uniquely as $(-1)^{a}5^{b}$, where $a \in \lbrace 1,0\rbrace$ and $0 \leq b <2^{n-2}$. Then you construct an isomorphism, sending $(-1)^{a}5^{b} \to (a,b) \in \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2^{n-2}\mathbf{Z}$
mich95
- 8,935
-
What is $l$ ? and how it follows that every element can be written uniquely as $(-1)^a5^b$ ? – Jun 01 '15 at 15:53
-
-
Ok, I will leave it to you to prove uniqueness (Assume $(-1)^{a}5^{b}=(-1)^{a'}5^{b'}$ mod $2^{l}$ and reduce mod 4). Now for existence, you have shown that 5 generates a subgroup of index 2 in $U_{\mathbf{Z}_{2}}$ and you can show that $-1$ does not belong to the subgroup generated by 5. It's a good exercise to see why it is enough! – mich95 Jun 01 '15 at 18:31
-
To show that $-1$ does not belong to the subgroup generated by $5$, I think one needs to use $5^{2^{n-3}}\equiv1+2^{n-1}\pmod{2^n}$, the proof of which requires induction. – ashpool May 31 '20 at 03:07
0
You prove that $5$ has order $2^{n-2}$ by induction.
lhf
- 221,500
-
See also http://math.stackexchange.com/questions/478082/mathbbz-2n-mathbbz-is-not-cyclic-group-for-n-geq-3. – lhf Jun 01 '15 at 14:37