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Let me check on this...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.

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- #36

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Let me check on this...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.

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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 $$\{1,-1\}$$Let me check on this...

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Your idea is correct, but your notation is incorrect.⋅⋅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##...

You can state that in ##\mathbb{Z_9}## , the Ring of Integers Modulo 9: ##~ [6]⋅[3]=[18]=[0]## . This is written in terms of the residue classes mod 9., which are the actual elements of the ring.

Added in **Edit**:

I should have also explicitly enumerated the elements (members) of this ring.They are: [0], [1], [2], [3], [4], [5], [6], [7], [8] .

Thus ##\displaystyle \mathbb{Z_9}=\left\{[0],\,[1],\,[2],\,[3],\,[4],\,[5],\,[6],\,[7],\,[8]\right\}## .

end of **Edit**.

Or as a congruence, you can state that ##~ 6⋅3\equiv 0 \mod 9## .

Or with modular arithmetic: ##~ 6⋅3=0 ## modulo ##9## .

Using the mod operation (a function) you could write: ##~~ (6⋅3) \!\!\!\mod9=0##.

For LaTeX: when writing mod, precede it with a back-slash as in \mod .

##2⋅5=1\mod9## gives ##2⋅5=1\mod9##

##2⋅5=1mod9## gives ##2⋅5=1mod9##

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- #39

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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;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 $$\{1,-1\}$$

http://campbell.mcs.st-and.ac.uk/~john/MT4521/Lectures/L3.html

Now on the complex number ##e^{iθ}## this to my understanding is another representation of ##cos θ + i sin θ##. They basically used the conversion from the cartesian coordinates to polar form. We know that ##e^{iθ}=cos θ + i sin θ##. (Euler's equation)

Now on the second part where they indicate that the rotation in a unit circle has determinant of ##1##...this is dependant (in my thinking) on the direction of rotation;

if the direction is say ##90^0## anticlockwise then our matrix of rotation would be;

##\begin{pmatrix}

1& 0\\

0 &1\end {pmatrix}##

if the rotation is say ##270^0## anticlockwise then our rotation matrix would be;

##\begin{pmatrix}

1& 0\\

0 &-1\end {pmatrix}##

giving us determinant values of ##1## and ##-1## respectively. I hope i am getting this right.

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- #40

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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).

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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##.

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).

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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≠0 mod 9## .

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.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.

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

Similarly, the

- #45

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Thanks @SammyS noted...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 isthe integer ##1##.not

Similarly, thezeroof this ring is ##[0]##. This isthe integer ##0##.not

- #46

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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 isthe integer ##1##.not

Similarly, thezeroof this ring is ##[0]##. This isthe integer ##0##.not

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;

##5\mod 2≡7## when ##k=1##.

##5\mod 2≡9## when ##k=2##.

##5\mod 2≡11## when ##k=3##...

- #47

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Technically , for the above ##k=-2##, although that's fine.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;

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.##5\mod 2≡7## when ##k=1##.

##5\mod 2≡9## when ##k=2##.

##5\mod 2≡11## when ##k=3##...

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##.

- #48

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This ought to be pretty obvious to me i.e The Euclidean algorithm...sometimes my thinking is not straight,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##.

anyway,

i note that if;

##7=2\cdot3+1## then we may either have;

##7\mod2=1## or

##7\mod3=1##

In general,

##a=b⋅c+d## then it follows that;

##a\mod c=d##

##a\mod b=d## dependant on either ##b## or ##c## being the quotient and divisor respectively.

- #49

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True, but in the present discussion, the context was ##\mod 2## .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##

Thus ##2## is the divisor, ##3## is the quotient, and the remainder is ##1## .

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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.

I can see that under Z/7Z; we have inverses that are commutative under multiplication i.e

##2⋅4=1##

##3⋅5=1##

##1⋅1=1##

##6⋅6=1## only

- #51

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Am still looking at the literature, i need confirmation on ##6##;

Now we know that;

My understanding- In reference to 6;

Let ##r=3## and ##p=2##, then it follows that,

##3^2+3^2=18\mod2=0##

##(3+3)^2=36\mod2=0##

satisfies definition 2.3.1 (1)

##3^2⋅ 3^2=81\mod2=1##

##(3⋅3)^2=81\mod2=1##

satisfies definition 2.3.1 (2)

using another example say, ##r=2## and ##p=3##, then it follows that,

##2^3+2^3=16\mod3=1##

##(2+2)^3=64\mod3=1##

satisfies definition 2.3.1 (1)

##2^3⋅ 2^3=64\mod3=1##

##(2⋅2)^3=64\mod3=1##

satisfies definition 2.3.1 (2)

thus the pth power map is a ring homomorphism.

Now we know that;

My understanding- In reference to 6;

Let ##r=3## and ##p=2##, then it follows that,

##3^2+3^2=18\mod2=0##

##(3+3)^2=36\mod2=0##

satisfies definition 2.3.1 (1)

##3^2⋅ 3^2=81\mod2=1##

##(3⋅3)^2=81\mod2=1##

satisfies definition 2.3.1 (2)

using another example say, ##r=2## and ##p=3##, then it follows that,

##2^3+2^3=16\mod3=1##

##(2+2)^3=64\mod3=1##

satisfies definition 2.3.1 (1)

##2^3⋅ 2^3=64\mod3=1##

##(2⋅2)^3=64\mod3=1##

satisfies definition 2.3.1 (2)

thus the pth power map is a ring homomorphism.

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- #52

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Are 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##.

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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.

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Thanks... we can't have ##1^p=1## because ##p## has to be prime ... unless I am missing something here.

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.

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What do you mean by "both ##n## values" being "units" ?Also here; i need insight on the inverse of a matrix...

View attachment 301452

Are 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##.

As @Maarten Havinga states, they are referring to the

Show that the map given as ##\psi## is the inverse (as in function inverse) of the map given as ##\phi## .

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- #56

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What do you mean by "both ##n## values" being "units" ? ...##n=1##.What do you mean by "both ##n## values" being "units" ?

As @Maarten Havinga states, they are referring to theinverse map.

Show that the map given as ##\psi## is the inverse (as in function inverse) of the map given as ##\phi## .

Now understood.

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