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Dummit and Foote - Definition of Zn - Motivation???

Peter

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MHB Site Helper
Jun 22, 2012
2,918
On page 56 (see attachment) Dummit and Foote define the notation \(\displaystyle Z_n \) as follows:

"Notation: For each \(\displaystyle n \in \mathbb{Z}^+ \), let \(\displaystyle Z_n \) be the cyclic group of order n (written multiplicatively). " (my emphasis)

But this notation is surely a bit counter-intuitive since \(\displaystyle Z_n \) is an additive group.

Indeed, Gallian defines/explains \(\displaystyle Z_n \) in EXAMPLE 2, page 74, as follows:

EXAMPLE 2 The set \(\displaystyle Z_n = \{ 0,1, 2, ... \ ... ,n-1 \} \) is a cyclic group under addition modulo n.


Surely the Gallian notation and explanation of the group is clearer.


The D&F definition/notation leads to the need for constant vigilance as in the example in Chapter 3, page 74, illustrating quotient groups where \(\displaystyle Z_n \) is defined as \(\displaystyle <x> \) with elements \(\displaystyle x^a \) - but of course the elements are actually of the form \(\displaystyle x + x + ... \ ... + x \) (a terms). Then we read statements like (see attachment - top of page 75)

"The multiplication in \(\displaystyle Z_n \) is just \(\displaystyle x^a x^b = x^{a+b} \)"

Why use "multiplication" to describe an operation that is actually addition?

Surely this is not intuitive - nor is it pedagogically helpful.

What do forum members think?

Can someone explain the likely motivation for D&F adopting this notation? What are the advantages of such a notation?

Peter
 

mathbalarka

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MHB Math Helper
Mar 22, 2013
573
You are quite right, the multiplicative notation is needed some getting used to since what we do over cyclic groups (in additive sense) is really addition modulo n. Nevertheless, one mustn't forget that notation doesn't really matter, since addition modulo n and usual addition is something different.

As for why Dummit & Foote uses it, I believe it's just because we often use multiplication rather than addition in several groups, just that.

I know it is difficult to get through that chapter just because the notation, I have read it, and the same problem occurred to me. The best one can do is to translate everything into a personal notebook from multiplicative to additive sense -- that's what I have done, at least.
 

Deveno

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MHB Math Scholar
Feb 15, 2012
1,967
When speaking of groups, what is often meant is the isomorphism class of a particular group (isomorphism is an equivalence relation on the (proper) class of all groups-the collection of all groups is "too big" to be called a set).

One therefore speaks of "the" cyclic group of order $n$, even though there are many possible "instantiations" (instances in a concrete sense) of a particular abstract group.

Abstractly, a cyclic group of order $n$ is ANY free group on one letter (pick any symbol you like for this letter...any one-element set will do, I will use $\{x\}$) subject to the one relation (the normal subgroup generated by the relator: $x^n = xx\cdots x$), $x^n = xx\cdots x = [\ \ ]$ (the "empty word", which is the identity of the free group).

It is common-place to identify this free group with the integers like so:

$x \leftrightarrow 1$
$[\ \ ] \leftrightarrow 0$ (0 is just a "placeholder")
$xx \leftrightarrow 1+1$ (addition is "concatenation of ones").

Thus:

$F(\{x\}) \cong \Bbb Z$
$\langle x^n \rangle \cong n\Bbb Z$
$F(\{x\})/\langle x^n\rangle \cong \Bbb Z/n\Bbb Z$

If one writes this quotient of a free group simply as:

$G = \{e,a,\dots,a^{n-1}\}$ with $a^n = e$ this isomorphism yields the correspondences:

$a \leftrightarrow [1]$ (the congruence class of 1 modulo $n$)
$a^k \leftrightarrow k[1] = [k]$
$e \leftrightarrow [0]$
$a^{-1} \leftrightarrow -[1] = [-1] = [n-1]$
$(a^k)(a^m) = a^{k+m (\text{mod }n)} \leftrightarrow [k] + [m] = [k+m]$

One should realize that the "particular" form an abstract group takes does not affect its algebraic properties, which are phrased entirely in terms of some set and a group operation. Whether we write a group additively or multiplicatively is purely a matter of NOTATION. Conventionally, it is usual to write abelian groups additively (considering them as $\Bbb Z$-modules), the usual module axioms for an abelian group are "snuck in" when writing a group multiplicatively as "rules of exponents":

$(a^m)^n = a^{mn}$
$a^{m+n} = (a^m)(a^n)$
$(ab)^m = (a^m)(b^m)$<--this one ONLY works for ABELIAN groups.
$a^1 = a$

Compare these to the module axioms:

$n(ma) = (nm)a$ (which equals $(mn)a$ since multiplication is commutative in $\Bbb Z$)
$(m+n)a = ma + na$
$m(a+b) = ma + mb$
$1a = a$

So.....why would we EVER want to write a cyclic group multiplicatively?

Well, consider the RING $\Bbb Z_p$ under addition and multiplication modulo $p$, a prime. It turns out that $(\Bbb Z_p)^{\times}$, the group of multiplicative units, is cyclic. To avoid confusion with the underlying ADDITIVE group $(\Bbb Z_p,+)$, it is usual to regard the group of units as $\langle b \rangle$, where $b$ is a primitive element (an element of multiplicative order $p-1$), written multiplicatively.

Also, multiplication is often written simply as concatenation:

$a \ast b \equiv ab$

Which means we can use considerably fewer symbols in writing out elements of a group.

When encountering a group, one needs to distinguish between its isomorphism class, and "which (particular) group it is". For example:

The cyclic group of order 4 can be realized as:

$\Bbb Z_4 = \{[0],[1],[2],[3]\}$ (often the brackets are omitted, and understood from context), under addition modulo 4.

$(\Bbb Z_5)^{\times} = \{[1],[2],[4],[3]\}$, under multiplication modulo 5 (here [4], NOT [2], is the unique element of order 2, and [2] or [3] are the generators, not [1] or [3]).

$\{1,i,-1,-i\}$, under complex multiplication.

$\{e,a,a^2,a^3: a^4 = e\}$ under "abstract" multiplication.

$\left\{\begin{bmatrix}1&0\\0&1 \end{bmatrix},\begin{bmatrix}0&-1\\1&0 \end{bmatrix},\begin{bmatrix}-1&0\\0&-1 \end{bmatrix},\begin{bmatrix}0&1\\-1&0 \end{bmatrix}\right\}$ under matrix multiplication.

The subgroup $\langle r \rangle$ of the dihedral group $D_4$ of order 8.

The subgroup $\{e, (1\ 2\ 3\ 4), (1\ 3)(2\ 4), (1\ 4\ 3\ 2)\}$ of the permutation group $S_4$ under permutation composition.

All of these are "different sets" with "different operations", and yet the same algebraic rules apply to all: there is one element of order 1 (the identity), one element of order 2, two elements (generators) of order 4. All these groups are abelian, and all have a unique subgroup of order 2 (the only non-trivial divisor of 4).

Talking about the isomorphism class allows us to handle all these cases at once, without re-deriving these facts anew for each instance. Now, ask yourself...what notation makes the most sense for this isomorphism class?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
When speaking of groups, what is often meant is the isomorphism class of a particular group (isomorphism is an equivalence relation on the (proper) class of all groups-the collection of all groups is "too big" to be called a set).

One therefore speaks of "the" cyclic group of order $n$, even though there are many possible "instantiations" (instances in a concrete sense) of a particular abstract group.

Abstractly, a cyclic group of order $n$ is ANY free group on one letter (pick any symbol you like for this letter...any one-element set will do, I will use $\{x\}$) subject to the one relation (the normal subgroup generated by the relator: $x^n = xx\cdots x$), $x^n = xx\cdots x = [\ \ ]$ (the "empty word", which is the identity of the free group).

It is common-place to identify this free group with the integers like so:

$x \leftrightarrow 1$
$[\ \ ] \leftrightarrow 0$ (0 is just a "placeholder")
$xx \leftrightarrow 1+1$ (addition is "concatenation of ones").

Thus:

$F(\{x\}) \cong \Bbb Z$
$\langle x^n \rangle \cong n\Bbb Z$
$F(\{x\})/\langle x^n\rangle \cong \Bbb Z/n\Bbb Z$

If one writes this quotient of a free group simply as:

$G = \{e,a,\dots,a^{n-1}\}$ with $a^n = e$ this isomorphism yields the correspondences:

$a \leftrightarrow [1]$ (the congruence class of 1 modulo $n$)
$a^k \leftrightarrow k[1] = [k]$
$e \leftrightarrow [0]$
$a^{-1} \leftrightarrow -[1] = [-1] = [n-1]$
$(a^k)(a^m) = a^{k+m (\text{mod }n)} \leftrightarrow [k] + [m] = [k+m]$

One should realize that the "particular" form an abstract group takes does not affect its algebraic properties, which are phrased entirely in terms of some set and a group operation. Whether we write a group additively or multiplicatively is purely a matter of NOTATION. Conventionally, it is usual to write abelian groups additively (considering them as $\Bbb Z$-modules), the usual module axioms for an abelian group are "snuck in" when writing a group multiplicatively as "rules of exponents":

$(a^m)^n = a^{mn}$
$a^{m+n} = (a^m)(a^n)$
$(ab)^m = (a^m)(b^m)$<--this one ONLY works for ABELIAN groups.
$a^1 = a$

Compare these to the module axioms:

$n(ma) = (nm)a$ (which equals $(mn)a$ since multiplication is commutative in $\Bbb Z$)
$(m+n)a = ma + na$
$m(a+b) = ma + mb$
$1a = a$

So.....why would we EVER want to write a cyclic group multiplicatively?

Well, consider the RING $\Bbb Z_p$ under addition and multiplication modulo $p$, a prime. It turns out that $(\Bbb Z_p)^{\times}$, the group of multiplicative units, is cyclic. To avoid confusion with the underlying ADDITIVE group $(\Bbb Z_p,+)$, it is usual to regard the group of units as $\langle b \rangle$, where $b$ is a primitive element (an element of multiplicative order $p-1$), written multiplicatively.

Also, multiplication is often written simply as concatenation:

$a \ast b \equiv ab$

Which means we can use considerably fewer symbols in writing out elements of a group.

When encountering a group, one needs to distinguish between its isomorphism class, and "which (particular) group it is". For example:

The cyclic group of order 4 can be realized as:

$\Bbb Z_4 = \{[0],[1],[2],[3]\}$ (often the brackets are omitted, and understood from context), under addition modulo 4.

$(\Bbb Z_5)^{\times} = \{[1],[2],[4],[3]\}$, under multiplication modulo 5 (here [4], NOT [2], is the unique element of order 2, and [2] or [3] are the generators, not [1] or [3]).

$\{1,i,-1,-i\}$, under complex multiplication.

$\{e,a,a^2,a^3: a^4 = e\}$ under "abstract" multiplication.

$\left\{\begin{bmatrix}1&0\\0&1 \end{bmatrix},\begin{bmatrix}0&-1\\1&0 \end{bmatrix},\begin{bmatrix}-1&0\\0&-1 \end{bmatrix},\begin{bmatrix}0&1\\-1&0 \end{bmatrix}\right\}$ under matrix multiplication.

The subgroup $\langle r \rangle$ of the dihedral group $D_4$ of order 8.

The subgroup $\{e, (1\ 2\ 3\ 4), (1\ 3)(2\ 4), (1\ 4\ 3\ 2)\}$ of the permutation group $S_4$ under permutation composition.

All of these are "different sets" with "different operations", and yet the same algebraic rules apply to all: there is one element of order 1 (the identity), one element of order 2, two elements (generators) of order 4. All these groups are abelian, and all have a unique subgroup of order 2 (the only non-trivial divisor of 4).

Talking about the isomorphism class allows us to handle all these cases at once, without re-deriving these facts anew for each instance. Now, ask yourself...what notation makes the most sense for this isomorphism class?
Thanks Deveno.

Working as I do without University tutors or professors, such a post is tremendously helpful.

When you state that \(\displaystyle Z_4 = \{[0],[1],[2],[3]\} \) and include the brackets indicating cosets or equivalence classes, you raise a question I have worried over - whether \(\displaystyle Z_n \) and \(\displaystyle Z/nZ \) are simply the same group or are isomorphic (subtle difference between the same and isomorphic, I know, but there is, I think, a difference.

If we write \(\displaystyle Z_4 = \{[0],[1],[2],[3]\} \) and indicate the elements as cosets then I think this group is not just isomorphic to \(\displaystyle Z/4Z \), but it is the same.

[Unless of course, we view the sets of elements of \(\displaystyle Z_4 \) and \(\displaystyle Z/4Z \) as the same but the 'difference' is the operation - modulo 4 arithmetic in one case and coset addition in the other?]

I was rationalising the difference this way - \(\displaystyle Z_4 \) consisted of the numbers 0, 1, 2, 3 with the operation being arithmetic in Z modulo 4 while \(\displaystyle Z/4Z \) was the group of cosets [0],[1],[2],[3] with the group operation being the addition of cosets - so they are not the same group but are isomorphic.

Can you please comment and clarify this matter?

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
That's not a bad way to think of it: one can think of $\Bbb Z_4$ as just consisting of "elements" 0,1,2, and 3, which only share the same SYMBOLS as the corresponding integers. In this view, the "modding" is taken care of by the OPERATION, which is thus a DIFFERENT operation that "ordinary" addition. That is, if one views these as "clock numbers" then "0" doesn't mean the INTEGER 0, but rather the "0-th position" (straight up), and (working clockwise) "1" just means (left), "2" means (straight down), and "3" means (right). Thus there IS no "4", we just have: 1 + 3 = 0.

However, the beauty of modular arithmetic (and why it is preferable to use this view instead of the view above), is that one can "do the modding" pre- or post-operation. This is what is meant by:

$[k] + [m] = [k+m]$.

For example, to evaluate:

$11 + 32$ (mod 7)

it makes no difference if we do this:

$[11] + [32] = [4] + [4] = [8] = [1]$

or:

$[11] + [32] = [43] = [1]$

observing that 8 = 7(1) + 1, and 43 = 7(6) + 1.

The first way is obviously going to be more convenient when k and m are large to begin with (this becomes especially apparent when one is doing multiplication mod n).

Normally, however, $\Bbb Z_n$ is used interchangeably with $\Bbb Z/n\Bbb Z$, and the former group you speak of is usually denoted $C_n$ (and written multiplicatively).

The question you pose is essentially an ontological one: what ARE actual groups (really)? Unfortunately, there isn't a truly definitive answer for this: mathematicians themselves do not often agree on just exactly what mathematical objects ARE. There are some (called Platonists) who hold that mathematical objects (such as the integer 2) have a "real" existence as ideal objects of thought, while others (called Fictionalists) hold that these are just imaginary objects that play a role in a game we invented called mathematics.

Still others (called Structuralists) hold that the ontological status of mathematical objects is irrelevant, as long as they have the desired properties. Most working mathematicians are non-Platonists while they're philosophizing and Platonists while they're working.

The problem is much deeper than you probably realize: the current "foundations" of mathematics (for lack of a more widely-accepted substitute) is to see everything in terms of sets (and functions, which are a special KIND of set). The trouble is, while an acceptable axiomatic foundation of sets exists (at least, the prevailing attitude is that if it were logically inconsistent, we would have known by now), the ontological question of what, exactly a set IS (besides the default answer of: something that behaves according to the set axioms) has not been answered. It is not even known if the entire universe could be considered a model for set theory (it probably cannot, because it is most likely finite, but even this is not known).

In any case, as long as you are JUST doing group theory, it doesn't matter: the GROUP properties of two isomorphic groups are the same. If you are working in some larger context, then it might matter WHICH group you have (for example, a group may have several isomorphic subgroups, and you want to count them...now you are no longer working entirely in group theory, but also in lattice theory as well).

Different text authors are going to handle these issues in different ways. The best advice I can give you is to pay close attention to the CONDITIONS listed for any given theorem, and to consider carefully the CONTEXT. Statements which appear to be "dissimilar" may well be expressing the same idea in a different garb.