Understanding the operation in ##(\mathbb{z_6})^{*}##

[chwala]

TL;DR

Kindly see attached notes

...Out of interest am trying to go through the attached notes,

My interest is on the highlighted, i know that in

$\mathbb{z}/\mathbb{6z}$ under multiplication we shall have:

$1*1=1$
$5*5=1$ am assuming that how they have the $(\mathbb{z}/\mathbb{6z})^{*}={1,5}$ is that correct?

....

In $\mathbb{z}/\mathbb{5z}$ we shall have,
$1*1=1$
$2*3=1$
$3*2=1$
$4*4=1$ .....integral domain...

insight or guidance is welcome...

Now the part i am not getting is the highlighted part in red.
Cheers.

 

[fresh_42]

The star in the notation means, the (multiplicative) group of units (invertible elements) in the ring. In the case of $\mathbb{Z}_6,$ we have only two units, $1$ and $5$, because there is no solution to the equations $a\cdot x=1$ if $a\in \{0,2,3,4\}.$ You are right, $1\cdot 1= 1$ and $5\cdot 5=1.$ There is only one group with two elements, so $\mathbb{Z}^*_6 =\{1,5\}\cong \mathbb{Z}_2.$

$\mathbb{Z}_5$ on the other hand is a field because $5$ is a prime number. As a field, all elements except $0$ are units, i.e. $\mathbb{Z}^*_5=\{1,2,3,4\}.$ You have found all the inverse elements. The words you highlighted in red simply mean that the number of elements in $\mathbb{Z}^*_5=\{1,2,3,4\}$ is four: $\# \mathbb{Z}^*_5=\# \{1,2,3,4\}=|\{1,2,3,4\}|=4.$

 

[chwala]

The star in the notation means, the (multiplicative) group of units (invertible elements) in the ring. In the case of $\mathbb{Z}_6,$ we have only two units, $1$ and $5$, because there is no solution to the equations $a\cdot x=1$ if $a\in \{0,2,3,4\}.$ You are right, $1\cdot 1= 1$ and $5\cdot 5=1.$ There is only one group with two elements, so $\mathbb{Z}^*_6 =\{1,5\}\cong \mathbb{Z}_2.$

$\mathbb{Z}_5$ on the other hand is a field because $5$ is a prime number. As a field, all elements except $0$ are units, i.e. $\mathbb{Z}^*_5=\{1,2,3,4\}.$ You have found all the inverse elements. The words you highlighted in red simply mean that the number of elements in $\mathbb{Z}^*_5=\{1,2,3,4\}$ is four: $\# \mathbb{Z}^*_5=\# \{1,2,3,4\}=|\{1,2,3,4\}|=4.$

...so in simple terms for a set to be defined as a field,it means that every element in that set must have an inverse like in the case of $\mathbb{z}_5?$. In $\mathbb{z}_6$ the elements $[2,3,4]$ have no inverse, thus not a field... or in other terms a lack of zero divisors...need to review the definitions...

lastly, i want to know if i am getting this right,

$\mathbb{z}_2 ×\mathbb{z}_3=(0,1)×(0,1,2)=(0,0), (0,1),(0,2),(1,0),(1,1)$ and $(1,2)$

what about $\mathbb{z}_2 ×\mathbb{z}_2 ×\mathbb{z}_2?$ let me give it a try a minute...

 

[fresh_42]

...so in simple terms for a set

ring

to be defined as a field,it means that every element in that set

except $0$

must have an inverse like in the case of $\mathbb{z}_5?$.

This is a necessary condition, not a sufficient condition.

We were talking about rings of the form
$$
\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n =\{0,1,\ldots,n-1\}.
$$
Bézout's lemma says that the greatest common divisor of $n$ and $k$ can be written as
$$
\operatorname{gcd}(k,n) = a\cdot k + b\cdot n
$$
If $\operatorname{gcd}(k,n)=1$ then $1\equiv a\cdot k \pmod{n}$ and $k$ has an inverse $a.$ However, if $n=p$ is prime, then all non-zero elements in $\left(\mathbb{Z}/p\mathbb{Z}\right)^* =\mathbb{Z}_p^*=\{1,\ldots,p-1\}$ are coprime to $p$ so they all have an inverse. Since all other properties of a field are already true in the ring $\mathbb{Z}_p^*,$ we have a field.

If $\operatorname{gcd}(k,n)=d>1$ then $d\,|\,n$ and we have $n=d\cdot e$ or $d\cdot e\equiv 0\pmod{n}.$ This means, we have a zero divisor $d\neq 0$ which is not allowed in a field, $d$ has no (multiplicatively) inverse element.

Both statements together say that $\mathbb{Z}/n\mathbb{Z}= \mathbb{Z}_n$ is a field if and only if $n=p$ is prime.

 

[chwala]

aaaaaaah now becoming clear...getting this concepts requires patience...i now know that,

$\mathbb{z}_2 ×\mathbb{z}_3=(0,1)×(0,1,2)=(0,0), (0,1),(0,2),(1,0),(1,1)$ and $(1,2)$ and on taking successive multiples of $(1,1)$ we get,

$(0,1),(1,1),(0,0),(0,2),(1,0)$ and $(1,2)$

since we can get the whole group by taking multiples of $(1,1)$ it follows that $\mathbb{z}_2 ×\mathbb{z}_3$ is cyclic of order 6. Bingo!

 

[fresh_42]

aaaaaaah now becoming clear...getting this concepts requires patience...i now know that,

$\mathbb{z}_2 ×\mathbb{z}_3=(0,1)×(0,1,2)=(0,0), (0,1),(0,2),(1,0),(1,1)$ and $(1,2)$ and on taking successive multiples of $(1,1)$ we get,

$(0,1),(1,1),(0,0),(0,2),(1,0)$ and $(1,2)$

since we can get the whole group by taking multiples of $(1,1)$ it follows that $\mathbb{z}_2 ×\mathbb{z}_3$ is cyclic of order 6. Bingo!

The other way around. Cyclic means, generated by a single element. There are two groups with six elements. One is $\mathbb{Z}_6=\mathbb{Z}_2\times \mathbb{Z}_3$ which is cyclic of order six, i.e. can be generated the way you described. $\mathbb{Z}_6$ is a direct product.

The second one is the symmetric group of three elements (permutations), $S_3=\mathbb{Z}_2\ltimes \mathbb{Z}_3=\mathbb{Z}_2\ltimes A_3.$ This group is not cyclic. It needs at least two elements to generate the entire group, e.g. the permutations $(1,2,3)$ and $(1,2).$ One generates the subgroup $\mathbb{Z}_2$ and the other one the normal subgroup $A_3,$ the alternating group with three elements. $S_3$ is a semidirect product.

 

[chwala]

it took me time to get to know what is happening in the cayleigh table.

just realised that the operation is addition ...time for break now. ...all the elements in the leading diagonal are all $(0,0)$...