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

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

1690463602101.png


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.
 
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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.##
 
fresh_42 said:
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...
 
Last edited:
chwala said:
...so in simple terms for a set
ring
chwala said:
to be defined as a field,it means that every element in that set
except ##0##
chwala said:
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.
 
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!
 
chwala said:
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.
 
it took me time to get to know what is happening in the cayleigh table.

1690476135120.png
just realised that the operation is addition :biggrin: ...time for break now. ...all the elements in the leading diagonal are all ##(0,0)##...
 
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