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Question is to Prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not cyclic Group for $n\geq 3$.

Hint : Find two subgroups of order $2$.

I somehow feel that a cyclic group can not have two distinct groups of same order. but, I am not sure about the proof.

I have no idea how to proceed for this.

any hint would be appreciated.

Thank You.

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  • $\begingroup$ For a proof using a different approach, see math.stackexchange.com/questions/66043/…. $\endgroup$ – lhf Aug 28 '13 at 10:57
  • $\begingroup$ @lhf: I am sorry, I did not understand the connection. It would be helpful if you can give a hint how does $(a^2)^{k−2}≡1(mod2^k) \forall k≥3$ implies $(\mathbb{Z}/2^n\mathbb{Z})^*$ is not cyclic Group for n≥3.$ $\endgroup$ – user87543 Aug 28 '13 at 11:03
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    $\begingroup$ The order of $(\mathbb{Z}/2^n \mathbb{Z})^*$ is $2^{n-1}$ but that result proves that no element has this order. $\endgroup$ – lhf Aug 28 '13 at 11:08
  • $\begingroup$ Ok, Ok. I got it. Thank YOu. $\endgroup$ – user87543 Aug 28 '13 at 11:14
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    $\begingroup$ Hint: How many elements of order $2$ would it have if it was cyclic? Can you find more than that? $\endgroup$ – Tobias Kildetoft Oct 23 '13 at 8:59
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Follow this outline:

If you only want to prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not cyclic, it is enough to prove that no element can have order $2^{n-1}$. For a proof, see How to prove by induction that $a^{2^{k-2}} \equiv 1\pmod {2^k}$ for odd $a$?.

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  • $\begingroup$ This Hint seems much difficult. But, As you have explained, No odd Integer has order $2^{n-1}$ (For even, order is at most n). So, $(\mathbb{Z}/2^n \mathbb{Z})^*$ has no element of order $2^{n-1}$ So is not cyclic. $\endgroup$ – user87543 Aug 28 '13 at 11:26
  • $\begingroup$ This edited version is more clear :). I see that $5^r, -5^r$ has order 2 for $r=2^{n-3}$. $5^r.5^r=5^{2r}=5^{2.2^{n-3}}=5^{2^{n-2}}=1$ (as any odd integer has order $2^{n-2}$ in $(\mathbb{Z}/2^n \mathbb{Z})^*$). So, we have $\big< 5^r\big>=\{1,5^r\}$ and similarly,$\big< -5^r\big>=\{1,-5^r\}$ which are two different groups of order 2. :) $\endgroup$ – user87543 Aug 28 '13 at 11:33
  • $\begingroup$ Not directly related, but just a comment: In fact $o(3)=o(5)=2^{n-2}$. $\endgroup$ – tc1729 Nov 10 '13 at 2:43
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Making lhf's fine (+1) answer perhaps a bit more concrete. There are three subgroups of order two: $H_1=\{1,-1\}$, $H_2=\{1,2^{n-1}+1\}$ and $H_3=\{1,2^{n-1}-1\}$. The non-1 element in each subgroup has square $\equiv1\pmod{2^n}$ as expected.

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  • $\begingroup$ The subgroups $H_2$ and $H_3$ are the subgroups in lhf's second bullet. $\endgroup$ – Jyrki Lahtonen Aug 28 '13 at 19:38
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Here is a simple way to do this :

  1. $U(8) = \{[1],[3],[5],[7]\}$, and check that $$ [3]^2 = [5]^2 = [7]^2 = [1] $$ so $U(8)$ is not cyclic (it doesn't have an element of order $4 = |U(8)|$)

  2. Since $8 \mid 2^n$ for $n > 3$, we have a natural ring homomorphism $$ \mathbb{Z}/2^n\mathbb{Z} \to \mathbb{Z}/8\mathbb{Z}, \text{ given by } [x]_{2^n} \mapsto [x]_8 $$ which induces a surjective group homomorphism $$ U(2^n) \to U(8) $$ Since the quotient of a cyclic group must be cyclic, it follows that $U(2^n)$ cannot be cyclic for $n\geq 3$

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  • $\begingroup$ you may refer book ""AA classical introduction to modern number theory" by Kenneth Ireland, Michael Rosen $\endgroup$ – thanks Oct 23 '13 at 17:21
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Consider $\pm(2^{k-1}+1)$. What is the order of these two elements if $k\geqslant 3$? Note they are $\neq \pm 1$ if $k\geqslant 3$.

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