- #36
chwala
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Let me check on this...WWGD said:Actually, no, these are cyclic subgroups of the group of all rotations. This group of all rotations is uncountable and uncountable groups cannot be cyclic.
Let me check on this...WWGD said:Actually, no, these are cyclic subgroups of the group of all rotations. This group of all rotations is uncountable and uncountable groups cannot be cyclic.
And rotation group is a nice example of infinite groups having finite subgroups and having " torsion": Nonzero elements A with ## A^n =0##, which seems counterintuitive. Notice the ring of countable integers being ( very) finitely generated by $$\{1,-1\}$$chwala said:Let me check on this...
Your idea is correct, but your notation is incorrect.chwala said:⋅⋅Ok, i see your point we cannot pick any item in ##\left [a_0\right]##, in that case we ought to have;
##[6]⋅[3]= 0 mod 9##
##[3]⋅[6]= 0 mod 9##
##[6]⋅[6]= 0 mod 9##
I can also see that,
##[2]⋅[5]= 1 mod 9## implying that ##[2]## is an inverse of ##[5]##, same to ##7## and ##4##...
I am familiarising with literature to respond amicably to the question, i would like clarification on the highlighted and i also hope that i am looking at the 'correct' literature;WWGD said:And rotation group is a nice example of infinite groups having finite subgroups and having " torsion": Nonzero elements A with ## A^n =0##, which seems counterintuitive. Notice the ring of countable integers being ( very) finitely generated by $$\{1,-1\}$$
Nice to read this...i think i will try and go through the whole link then get more insight so as to deal with post ##27##.mathwonk said:To me, typical examples of zero divisors occur in product rings, like ZxZ, where (1,0) x (0,1) = (0,0).
Note also that although in Z/6Z, we have 2.3 = 6 = 0, in fact Z/6Z ≈ Z/2Z x Z/3Z, via the map sending 1 to (1,1), and that 2 and 3 in Z/6Z correspond under this isomorphism to (0,2) and (1,0).
Notice that if you use a back-slash ( \ ) with 'mod', the above sentence is rendered as follows:chwala said:If i understand your question we are looking for ##xy≡0 mod 9## with the condition that ##x≠0 mod 9##and ##y≠0 mod 9## .
It is improper to state things such as ##[a_0]##=##9## , ##[a_2]##=##2## , etc.Now we have,
##[a_0]##=##\{\ldots, -9,0,9,18,27 \ldots\}##
##[a_1]##=##\{\ldots, -8,1,10,19,28 \ldots\}##
if we pick ##[a_0]##=##9## and ##[a_1]##=##1##, then ##[a_0]⋅[a_1]=9⋅1=0 mod 9##
just a minute i check ##1## cannot be a zero divisor in any ring!
Second attempt.
##[a_0]##=##\{\ldots, -9,0,9,18,27 \ldots\}##
##[a_2]##=##\{\ldots, -7,2,11,20,29 \ldots\}##
if we pick ##[a_0]##=##9## and ##[a_2]##=##2##, then ##[a_0]⋅[a_2]=9⋅2=0 mod 9## therefore ##9## and ##2## are zero divisors.
Thanks @SammyS noted...SammyS said:Notice that if you use a back-slash ( \ ) with 'mod', the above sentence is rendered as follows:
If I understand your question we are looking for ##xy=0\mod 9## with the condition that ##x≠0\mod 9## and ##y\ne0 \mod 9## .
Now onto my primary reasons for making this reply ...
It is improper to state things such as ##[a_0]##=##9## , ##[a_2]##=##2## , etc.
Presumably, by what you have stated early in the quoted post (Post #17), ##a_0## is some integer multiple of ##9## .
Similarly ##a_2## is one of the integers in the set ##\{\dots\, -7,\,2,\,11,\,20,\, \dots \}## . In fact you can say that ##a_2 \in[2]## meaning that
##a_2=9k+2## for some integer, ##k## .
You can pick ##a_0## to be ##9## . That makes ##[a_0]=[9]## . This refers to a set, not an individual integer.
That leads me to mention another misconception you seem to have. There is nothing special about ##a_0,\,a_1,\,a_2,\,a_3,\ ## etc. That is not any sort of general or widely accepted notation. The author of one of the attachments in Post #2, was simply picking a member from one of the residue classes, and using an index (in a subscript) to indicate from which class the pick was made. The point being made was that any particular class can be labeled using any of its members.
One last caution:
Be careful when referring to the ##1## or the ##0## of a ring. Those are, generically speaking, the multiplicative identity and the additive identity, respectively.
For the ring ##\mathbb{Z}_9## , the "1" is ##[1]##, This is not the integer ##1##.
Similarly, the zero of this ring is ##[0]##. This is not the integer ##0##.
SammyS said:Notice that if you use a back-slash ( \ ) with 'mod', the above sentence is rendered as follows:
If I understand your question we are looking for ##xy=0\mod 9## with the condition that ##x≠0\mod 9## and ##y\ne0 \mod 9## .
Now onto my primary reasons for making this reply ...
It is improper to state things such as ##[a_0]##=##9## , ##[a_2]##=##2## , etc.
Presumably, by what you have stated early in the quoted post (Post #17), ##a_0## is some integer multiple of ##9## .
Similarly ##a_2## is one of the integers in the set ##\{\dots\, -7,\,2,\,11,\,20,\, \dots \}## . In fact you can say that ##a_2 \in[2]## meaning that
##a_2=9k+2## for some integer, ##k## .
You can pick ##a_0## to be ##9## . That makes ##[a_0]=[9]## . This refers to a set, not an individual integer.
That leads me to mention another misconception you seem to have. There is nothing special about ##a_0,\,a_1,\,a_2,\,a_3,\ ## etc. That is not any sort of general or widely accepted notation. The author of one of the attachments in Post #2, was simply picking a member from one of the residue classes, and using an index (in a subscript) to indicate from which class the pick was made. The point being made was that any particular class can be labeled using any of its members.
One last caution:
Be careful when referring to the ##1## or the ##0## of a ring. Those are, generically speaking, the multiplicative identity and the additive identity, respectively.
For the ring ##\mathbb{Z}_9## , the "1" is ##[1]##, This is not the integer ##1##.
Similarly, the zero of this ring is ##[0]##. This is not the integer ##0##.
Technically , for the above ##k=-2##, although that's fine.chwala said:This part is getting clear to me, in general,
if ##a=b\mod n## then it follows that ##a=b+kn## for some ##k∈ \mathbb{Z}##
For example, we know that
##1≡5\mod 2## then using ##1=5+2k## then the other equivalent values would be given as follows;
As mathematical function, ##5\mod2## gives the remainder for 5 divided by 2, in the sense of Euclidean Division. In other words, ##5\mod2=1## because ##5=2\cdot2+1## , 2 being the quotient, 1 being the remainder. A rather poor example.##5\mod 2≡7## when ##k=1##.
##5\mod 2≡9## when ##k=2##.
##5\mod 2≡11## when ##k=3##...
This ought to be pretty obvious to me i.e The Euclidean algorithm...sometimes my thinking is not straight,SammyS said:Technically , for the above ##k=-2##, although that's fine.
Also, usually the mod 2 is enclosed in parentheses as in: ##\quad 1≡5\ \ (\mod 2)##As mathematical function, ##5\mod2## gives the remainder for 5 divided by 2, in the sense of Euclidean Division. In other words, ##5\mod2=1## because ##5=2\cdot2+1## , 2 being the quotient, 1 being the remainder. A rather poor example.
Better: ##7\mod2=1## because ##7=2\cdot3+1## , 3 being the quotient, 1 being the remainder.
Now, if you want to write any of these as congruence mod 2 , then the ##\mod2## usually appears in parentheses at the tail end of the statement.
##5\equiv7\ \ (\mod 2 \ )## means ##7=5+2k\text{, where }k=1## .
A bit cleaner way to consider such a congruence is to see that the following is true.
##5\equiv7\ \ (\mod 2 \ )## means ##7-5=2k\ ## for some integer, ##k##.
True, but in the present discussion, the context was ##\mod 2## .chwala said:This ought to be pretty obvious to me i.e. The Euclidean algorithm...sometimes my thinking is not straight ,
anyway,
i note that if;
##7=2\cdot3+1## then we may either have;
##7\mod2=1## or
##7\mod3=1##
Hey, just confirm that i am getting you right, for Z/7Z we do not have zero divisors! For Z/6Z yes we do have zero divisors.mathwonk said:I guess it is even simpler when viewed algebraically, as quotient rings. I.e. in a quotient ring, something is set equal to zero, and it just depends on whether or not the thing you set equal to zero is or is not a prime. The ring of functions on the x-axis in the plane, can be viewed as the polynomial ring in X and Y where Y is set equal to zero, and Y is prime. The ring of functions on the union of x and y axes, is the polynomial ring in X and Y, but with the non prime product XY set equal to zero. The ring Z/6Z is the ring Z but with the non prime product 6 = 2.3 set equal to zero, while Z/7Z is Z with the prime 7 set equal to zero. When a product AB is set equal to zero, of course the elements A and B become zero divisors in the quotient ring.
Thanks... we can't have ##1^p=1## because ##p## has to be prime ... unless I am missing something here.Maarten Havinga said:The p-power is a homomorphism indeed based on your observations. The ##1^p=1## part is trivial but when handing something in do add it.
As to the matrix inverse, there's a slight confusion. They mean the inverse of the homomorphism as a map, not the inverse of the matrix. It's a rather trivial statement.
What do you mean by "both ##n## values" being "units" ?chwala said:Also here; i need insight on the inverse of a matrix...
View attachment 301452Are they assuming that in the given symmetric matrix; both ##n## values are units? What if we have non- unit ##n## values? ...then the inverse would not be equal to ##n##.
What do you mean by "both ##n## values" being "units" ? ...##n=1##.SammyS said:What do you mean by "both ##n## values" being "units" ?
As @Maarten Havinga states, they are referring to the inverse map .
Show that the map given as ##\psi## is the inverse (as in function inverse) of the map given as ##\phi## .