What the set $I$ of invertible elements in $\mathbb{Z}_{9}$?

Question

$5.2 = 1$

The set of invertible elements in $\mathbb{Z}_{9}$ is $I = \{1,2,5\}$.

Q1: Is it correct?

I was also asked to show that it is not isomorphic to $S_{3}$

How can I show that it is not isomorphic? Thank you.

Is the question $\Bbb{Z}_9$ not isomorphic to $\mathcal{S}_3$ or is it the multiplicative group of the invertible elements $\Bbb{Z}_9^{\times}$ that has six elements? — marwalix, Apr 17 '17 at 04:22
@marwalix Sorry Marwalix for answering your question late. The multiplicative group of the invertible elements is not isomorphic to $S_3$ — Zouhair El Yaagoubi, Apr 20 '17 at 10:52

DHMO · Accepted Answer · 2017-04-17T04:31:26.917

3

$$\begin{array}{c|c} \times&0&1&2&3&4&5&6&7&8\\\hline 0&0&0&0&0&0&0&0&0&0\\\hline 1&0&\color{red}1&2&3&4&5&6&7&8\\\hline 2&0&2&4&6&8&\color{red}1&3&5&7\\\hline 3&0&3&6&0&3&6&0&3&6\\\hline 4&0&4&8&3&7&2&6&\color{red}1&5\\\hline 5&0&5&\color{red}1&6&2&7&3&8&4\\\hline 6&0&6&3&0&6&3&0&6&3\\\hline 7&0&7&5&3&\color{red}1&8&6&4&2\\\hline 8&0&8&7&6&5&4&3&2&\color{red}1 \end{array}$$

Therefore, the invertible elements of $\Bbb Z_9$ are $\{1,2,4,5,7,8\}$.

Indeed, $S_3$ has $6$ elements while $\Bbb Z_9$ has $9$ elements, so they are not isomorphic to each other.

In general, for $\Bbb Z_n$, $x$ is invertible iff $\gcd(x,n)=1$.

By Bézout's Lemma, $\gcd(x,n)=1 \iff \exists a,b:ax+bn=1 \iff \exists a:ax=1$.

edited Apr 17 '17 at 04:31

answered Apr 17 '17 at 04:09

DHMO

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I think you mean $4, 7$ from your table? – q.Then Apr 17 '17 at 04:12
@q.Then Thanks, corrected – DHMO Apr 17 '17 at 04:12
@marwalix Ask the OP, not me. – DHMO Apr 17 '17 at 04:19
It's also probably worth noting somewhere that in general, the units in $\mathbb Z_n$ will be those elements $x$ such that gcd$(x,n)$=1. That is, precisely those elements coprime with the order $n$. – kholli Apr 17 '17 at 04:28
@kholli Is it better now? – DHMO Apr 17 '17 at 04:31

score 3 · Answer 2 · answered Apr 17 '17 at 04:10

Yes, you are correct. Clearly two groups that are not the same order cannot be isomorphic to each other...

We wish to find all elements that have a multiplicative inverse in $\mathbb{Z}_9$ (as indicated by your question). 9 is not prime, so some inverses may not exist, luckily for you, $\mathbb{Z}_9$ is a relatively small group to check. $$1, 2\cdot5, 4 \cdot 7, 8 \cdot 8$$

Those are the multiplicative inverse pairings

What the set $I$ of invertible elements in $\mathbb{Z}_{9}$?

2 Answers2