Is multiplicative mod Z9 cyclic?

4,893

Solution 1

The multiplicative groups of $\mathbb{Z}/9 \mathbb{Z}$ and $\mathbb{Z}/ 17\mathbb{Z}$ are indeed cyclic.

More generally, the multiplicative group of $\mathbb{Z}/p^k \mathbb{Z}$ is cyclic for any odd prime $p$.

If you are supposed to know this result, just invoke it. If you do not know this result, possibly you are expected to do this via a direct calculation.

To this end, you'd need to identify a generating element in each case. For example, for $9$ you have, trying $2$ as generator, $2^1= 2$, $2^2=4$, $2^2 =8=-1$, $2^4=-2$, $2^5=-4$, $2^6 = 1$.

Thus, $2$ indeed generates the multiplicative group of $\mathbb{Z}/9 \mathbb{Z}$, which has as its elements only the classes co-prime to $9$, that is, the six elements we got above.

However, the set $\mathbb{Z}/9 \mathbb{Z}$ (of nine elements) with multiplication, is not a group at all. For example, the class $0$ can never have a multiplicative inverse (neither have $3$ nor $6$).

Solution 2

Invertible elements of $\mathbb Z_m$ form a group which is cyclic only in the following cases

1) $m=2,4,8$

2) $m=p^k, k=1,2, 3,\ldots$ and $p$ is odd prime

3) $m=2p^k, k=1,2, 3,\ldots$ and $p$ is odd prime

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Idanhacm
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Idanhacm

Computer science student in Tel-Hai academic college.

Updated on August 13, 2022

Comments

  • Idanhacm
    Idanhacm about 1 year

    I'm asked if $\mathrm Z_9$or $\mathrm Z_{17}$ under multiplication is cyclic or not, proof needed. I understand the rules with addition, but under multiplication i can't make sens of the rules. I tried to manually calculate it with but did not find a generator thus i assumed that it is not cyclic but with $\mathrm Z_{17}$ it's a little bit more excruciating, what is the best way to prove if a group like this is cyclic or not? thanks in advance.

    • lulu
      lulu over 6 years
      $\mathbb Z_9$ is not a group under multiplication. $0$ is not invertible (for example).
    • quasi
      quasi over 6 years
      However, since $9$ and $17$ are both odd prime powers, in each of the rings $\mathbb{Z}_9$ and $\mathbb{Z}_{17}$, the multiplicative group of units is, in fact, cyclic, with orders $\phi(9)=6$ and $\phi(17)=16$, respectively.
    • Idanhacm
      Idanhacm over 6 years
      with what generator?
    • quasi
      quasi over 6 years
      Experiment -- start with a potential generator and take powers. The primitive element theorem guarantees success.
  • quid
    quid over 6 years
    It's not cyclic for $8$. Each of $3,5,7$ has order $2$ mod $8$.
  • Minz
    Minz about 6 years
    Oh, yes. My mind came to mind.