Proving multiplicative group is not cyclic

Asked Sep 11 '19 at 16:24

Active Sep 11 '19 at 16:28

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Prove that $\phi(2^n)$ for $n\geq 3$ is not a cyclic group.

$\phi(2^n)$ is a multiplicative group of integers modulo $2^n$.

Solution just for $n=3$:

$\phi(8)$ : $\{1,3,5,7\}$ (numbers that are co-prime with 8).

[3][5] = [7] mod8

[7][7] = [1] mod8

[3][3] = [1] mod8

$\phi(8)$ $\neq$ $<[3]>$

$\phi(8)$ $\neq$ $<[5]>$

$\phi(8)$ $\neq$ $<[7]>$

$\phi(8)$ $\neq$ $<[1]>$

Therefore, for $n=3$ this group is not cyclic.

I don't know how to prove that the group is not cyclic for $n \geq 3$.

edited Sep 11 '19 at 16:28

asked Sep 11 '19 at 16:24

1

Again, $f(2^n)$ is not really the best notation. You had this question a few days ago, with exactly the same notations. – Dietrich Burde Sep 11 '19 at 16:26
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I think $U(2^n)$ is one commonly accepted notation for this. – Arthur Sep 11 '19 at 16:27
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A common notation for this group is $(\Bbb Z / 2^n \Bbb Z)^*$. – Robert Shore Sep 11 '19 at 16:27

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