Chapter 1
Algebraic Systems
Definition 1.1. A binary operation on a set X is function f:XXX. The word binary refers
to the number of arguments of f.
Definition 1.2. An algebraic system or simply an algebra is a finite sequence (X,f1,f2,…,fn)
where X is a set and f1,f2, …, fn are (binary) operations on X.
Traditionally, we use symbols like +,*,-,\ to denote operations and we place them between the
arguments writing a+b rather than +(a,b). For any two arguments x and y, f(x,y) is called “the
result of f on x and y”.
The definition of an operation says that the result of the operation f for every two
arguments x and y from X belongs to X, we say then that X is closed with respect to the
operation f or closed under f. For example R is closed under addition, subtraction and
multiplication, which simply means that the sum, the difference and the product of any two
real numbers is a real number. The set N of natural numbers is not closed under subtraction
and is not closed under division either.
Sometimes it may be necessary to consider operations of other “arities” – unary (one
argument) operations, and in general n-ary (n-argument) operations.
An algebraic operation need not be anything as conventional as addition or
multiplication. In fact it may be absolutely any function, even apparently wild and
meaningless as long as it maps the product of X by itself into X.
Example 1.1. Consider the operation * defined on a set X as (x,yX) x*y=x. The definition
means that * assigns to every pair (x,y) the first element of the pair. It may not be very
exciting but it is a perfectly legal algebraic operation.
Example 1.2. This is another silly but formally correct example – the constant function that
assigns the same element of Z ,for example 7, to every pair (x,y) from ZZ, i.e. (x,yZ)
x&y=7.
Example 1.3. The following definition “whenever I am given a pair (x,y) I toss a coin and
chose x if it is heads, or y otherwise” is illegal because it may happen that presented for the
second time with the same pair (x,y) you will chose differently, so your procedure does not
define a function. You can correct it though by adding another condition, something like “If I
come across the same pair again, I stick to may previous choice”.
Example 1.4. Subtraction. It is an operation on the set of integers Z, but is not an operation
on the set of natural numbers N because in the case of X=N the result of subtraction does not
belong to X for some pairs of natural numbers (12=1).
And now let us do something really wild.
Example 1.5. Let ZOO={a cow, a dog, a frog}. We define an operation  on the set ZOO by
means of the operation table

cow dog frog
cow cow cow cow
dog dog dog dog
frog frog frog frog
This is another way of saying cowcow=cow, cowdog=cow, cowfrog=cow, dogcow=dog
and so on. Those poor darlings who ask now “But what does it mean a dog times a frog is a
dog?” are kindly requested to read this chapter again from the beginning, because it “means”
nothing (and that’s the whole fun). Some readers may notice, that our  operation is a special
case of the operation * from Example 1.1, namely this is what we get if we put X=ZOO.
Example 1.6. Let X=YY, i.e. X is the set of all functions mapping X into X. We will denote
by ◦ the operation of composition of functions. The result of this operation is a new function
f◦g from Y into Y, whose value on an element y from Y is defined as (f◦g)(y)=f(g(y)).
Example 1.7. Symmetric difference. For any set X, 2X denotes the set of all subsets of X. In
addition to the well-known set operations of union, intersection and set difference we will
consider the operation of the symmetric difference defined as AB=A\BB\A. Obviously, if
A and B belong to 2X then so do A\B and B\A and so does their union.
Example 1.8. Operations modulo n. First recall that k mod l denotes the remainder of the
division of k by l. Let n be a positive natural number. Consider the set Zn={0,1, … n-1}. For
any two numbers a and b from Zn we define their sum modulo n as anb=(a+b)modn.
Correspondingly, we define the product modulo n as anb=(ab)modn. By the definition of the
remainder of the division by n, both anb and anb are elements of Zn. Symbols n and n
are a nuisance so whenever there is no risk of ambiguity we shall simply write  or . We
must only remember that in (Z7,)  denotes addition mod 7 and in (Z5,) – mod 5. In these
finite algebras we have all sorts of funny identities. Things like “two plus two is zero” or
“three times three is one” are commonplace as long as you do your arithmetic “mod 4”.
To those readers, who have a problem with why are we using the  symbol for this
operation or * for that or & for yet another I can only say why not? You can switch the
symbols to your heart’s delight. It is not the symbols that count but what they symbolize.
Admittedly, some of the above examples make no real-life sense and it is hard to
imagine why would anybody want to study them. Why then we do study them? Let me put it
like this: in mathematics we often ask ourselves the question “what can we say about an
object if all we know is … (here comes a list of axioms)?”. At first we assume nothing about
the nature or the properties of our operation, but to assume nothing is to be prepare to
consider all, no matter how bizarre, operations. It is difficult to include the concept of
sensibility in the definition of an operation, so we just go ahead and formulate a catch-all
definition and filter out the nonsense later. Moreover, an operation that seems meaningless
now may turn out to be useful later – such things happen from time to time.
As our Example 1.3 shows, there may be no “formula” or “rule” for calculating the
result of an operation. You may think of an operation as a little black box. You push inside
two elements from your set X and the result just pops up. Some of these black boxes are
predictable and understandable (those with words like addition, multiplication and the like
printed on the top) while others are not. They are only bound by two rules: first, the result
must be from X, and second, if we push in a pair of objects for the second time, what pops up
must be the same as before.
Algebraic systems (algebras) are categorized according to the number of operations
and their properties. For instance we may consider algebras with one binary operation or with
two binary and one unary operation and so on. Properties of operations are laws like
associativity, commutativity and other, more exotic ones. Let us begin with associativity.
Definition 1.3. An operation f on X is said to be associative iff
(x,y,zX) f(x,f(y,z))=f(f(x,y),z)
Equivalently we may write (x,y,zX) xf(yfz)=(xfy)fz.
An algebra (X,∆) is said to be associative iff the operation ∆ is associative
The condition may look a bit strange at first, but let us try to see what it really means.
If you try to calculate the left hand side of the equality you have to apply f to x and the result
of f on y and z. On the other side you apply f to the pair consisting of the result of f on x and y
and of z. Notice that the order in which x,y and z appear on the left and on the right hand side
is the same. The only difference is in how we group them before we apply f. As an example
we consider first the operation of addition on the set of real numbers. Then the equality reads
+(x,+(y,z))=+(+(x,y),z), which can be translated into x+(y+z)=(x+y)+z and this is a very wellknown law of arithmetic; we say addition is associative. Now we will try to do the same trick
for subtraction. We get (x, (y,z)) = ((x,y),z), which, in the more familiar form, is
equivalent to x(yz)=(xy)z. As this evidently not true for all reals (e.g. x=y=0 and z=1)
we conclude by saying that subtraction is not associative. It is an easy exercise to see that the
operation * from Example 1.1 is associative.
The two mod n operations defined in Example 1.8 are associative. They are important
enough to deserve a special treatment.
Proposition 1.1. For each positive integer n, both (Zn,) and (Zn,) are associative algebras.
Proof. We will prove associativity of . Associativity of  can be proved in a similar way.
Let us first prove two simple facts
(1)
For every integer a (not necessarily from Zn) we have (a mod n)mod n = a mod n.
Proof of (1). Since a can be uniquely expressed as a=qn+r , 0  r  n1 and (qn+r)mod n = r,
and rmod n = r, we have (amod n)mod n = ((qn+r)mod n)mod n = rmod n = r = amod n.
(2)
For any two integers a and b (a+b)mod n = (amod n + bmod n)mod n.
Proof of (2). Let a=qan+ra , 0  ra  n1 and b=qbn+rb , 0  rb  n1. Hence
(a+b)mod n = (qan+ra+ qbn+rb)mod n = ((qa + qb)n + ra + rb)mod n = (ra + rb)mod n =
= (amod n + bmod n)mod n.
Now:
(ab)c = ((a+b)mod n+c)mod n
{by the definition of }
= (((a+b)mod n)mod n+cmod n)mod n
{thanks to (2)}
= ((a+b)mod n+cmod n)mod n
{because of (1)}
= ((a+b)+c)mod n
{we are using (2) again, only backwards}
= (a+(b+c))mod n
{because addition is associative}
= (amod n+(b+c)mod n) mod n
{we can thank (2) for this one}
= (amod n+((b+c)mod n)mod n) mod n
{it is (1) once more, only used backwards}
= (a+(b+c)mod n) mod n
{(2), bacjwards}
= a(bc)
{by the definition of }
Bingo! We have transformed (ab)c into a(bc).
Definition 1.4. An operation f on X is said to be commutative iff
(x,yX) f(x,y)=f(y,x)
Equivalently we may write (x,yX) xfy=yfx.
Consider the operation & on R defined as &(x,y)=x&y=x2+y. Remember that (x,y) is
an ordered pair of numbers, so the definition means “square the first element of the pair and
add the second one to the result”. Hence &(y,x)=y&x=y2+x and obviously the two are not
equal (not always equal, not for all x and y from R), which means that & is not commutative.
Addition and multiplication of real numbers are obviously commutative, which easily
implies that addition mod n and multiplication mod n are, too.
Subtraction and division (the last one defined on R\{0}) are obviously not
commutative.
Function composition in general is not commutative. For example take f(x)=x+5 and
g(x)=2x. Then f◦g(5)=(25)+5=15, while g◦f(5)=2(5+5)=20.
Definition 1.5. An algebra (X,f) is said to possess an identity element iff
(eX)(xX) f(x,e)=f(e,x)=x
Equivalently we may write (eX)(xX) xfe=efx=x. Each such e is called an
identity element of the operation f.
Proposition 1.2. If e and d are identity elements in an algebra (X,#) then e=d.
Proof. e#d=e because d is an identity and e#d=d#e=d because e is an identity. Since e and d
are both equal to e#d they are equal.
As a consequence, we can refer to e as the identity element of (X,#).
Identities for addition and multiplication are 0 and 1, respectively.
Example 1.9. The identity for function composition is the identity function id defined as
id(x)=x for every x from X. Indeed, for every xX (f◦id)(x) = f(id(x)) = f(x) which implies f◦id
= f. Similarly, (id◦f)(x) = id(f(x)) = f(x), hence id◦f = f.
Example 1.10. The identity for symmetric difference is  - the empty set. As for the other set
operations, the identity for  is obviously , while the identity for  is the universal set X.
Notice that the identity for the intersection varies from one set algebra to another. If we are
playing with subsets of R then R is the identity, if we are playing with subsets of X={a,b,c}
then the identity is {a,b,c} and so on.
Example 1.11. The identities for addition mod n and multiplication mod n ( and ) are 0
and 1, respectively.
Example 1.12. The operation x*y=x has no identity, and, consequently, the operation  from
our ZOO example has no identity.
Example 1.13. Subtraction has no identity. Admittedly, since x0=x, 0 is the right hand side
identity for subtraction but there is no “left” identity. The identity must be both sides identity,
even if the operation is not commutative, like function composition. Similarly division has no
identity.
Definition 1.6. Suppose e is the identity in an algebra (X,*). We say that b is an inverse
element for a iff a*b=b*a=e. We say then that a is invertible.
Proposition 1.3. Suppose (X,*) is an algebra, * is associative and (X,*) has the identity
element e. Then, for every aX, if a is invertible then its inverse element is unique. Formally
(a,b,cX)(a*b=b*a=e  a*c=c*a=e)b=c
Proof.
b= b*e = b*(a*c) = (b*a)*c = e*c = c 
Traditionally we denote the inverse to a by a-1. This causes no ambiguity as long as we
are dealing with a single operation algebra. In more complicated cases we will have to
indicate what kind of inverse element we are looking for. If we consider several operations on
the same set, a particular element may be invertible with respect to one and not invertible with
respect to the another. Also the inverse with respect to one operation may be different from
the inverse with respect to another. For example the inverse to 2 with respect to addition is –2,
with respect to multiplication it is ½ and there is no inverse to 2 with respect to subtraction,
because subtraction has no identity, so asking about an inverse with respect to subtraction is
meaningless.
Example 1.14. Every subset of X (every element of 2X) is its own inverse with respect to the
symmetric difference. This is clear as AA = A\AA\A =  =  for every set A. In
nontrivial cases subsets of X are not invertible with respect either to union or to intersection.
For example, AB is always a subset of A, so unless A=X, AB cannot be equal to X – the
identity for intersection.
Example 1.15. The case of function composition is more interesting in that some functions,
namely those that are 1-1 and “onto”, (they are known as bijections) are invertible and some
are not. The inverse element in case of one of those invertible functions is the inverse function
defined as f –1(x)=z iff f(z)=x, for all x and z from X. Indeed, we have f◦f1(x) = f(z) = x =
id(x) and f1◦f(z) = f1(x) = z = id(z), so f◦f1 = f1◦f = id. Why does it only work for
bijections? It is simple. If f is not “onto” then f1 is not defined for those x-s who are not
values of f. If f is not 11 then for some z and y, zy, f(z)=f(y)=x and f1(x)=z and f1(x)=y,
so f1 is not a function. There is another interesting point about inverse functions. If we look
at the definition of the inverse element b to some a we notice that it is symmetric in that b is
inverse to a iff a is inverse to b. This implies that the inverse to some element is itself
invertible. Hence the inverse function to a bijection is itself a bijection, and we don’t have to
prove that it is 11 and “onto”. It simply follows from general algebraic principles.
Example 1.16. In (Zn,), the inverse to k is (nk)mod n. Notice that 0 is its own inverse with
respect to addition modulo n. In (Zn,) the problem is much more complicated and will be
discussed later. Let us just mention here that in (Z8,) 3 is its own inverse (indeed, 33 =
(3·3)mod 8 = 9 mod 8 = 1), while 2 is not invertible – this can be verified by testing all
candidates for 2-1 and rejecting them one by one.
Definition 1.7. In an algebra (X,#,*) with two binary operations one may be distributive with
respect to the other. We say that * is distributive with respect to # iff for every a,b,c from X
we have a*(b#c) = (a*b)#(a*c) and (b#c)*a = (b*a)#(c*a)
As we all know, multiplication is distributive with respect to addition, a(b+c)=ab+ac
but addition is not distributive with respect to multiplication, for example 2+(3·4)=14 and
(2+3)·(2+4)=30.
In the set algebra (2X,,) we have both distributivities A(BC) = (AB)(AC)
and A(BC) = (AB)(AC). These laws follow from the logic laws of distributivity of
conjunction with respect to disjunction and vice versa.
One of the central concepts of the theory of algebraic systems is that of an
isomorphism between two algebras. For simplicity we will only consider binary algebras (i.e.
algebras with binary operations), but the definition can be easily extended to more general
cases.
Definition 1.8. An algebra (X,f1,f2, … ,fk) is said to be isomorphic to (Y,g1,g2, … ,gn) iff k=n
and there exists a bijection (i.e. a 11 and onto mapping) :XY such that for each i=1,…,n
and for each a,b from X
(fi(a,b)) = gi((a),(b))
Each such mapping is called an isomorphism.
In the traditional notation we write (a fi b) = (a) gi (b) and we say that 
preserves the operation fi.
Example 1.17. (R+,·) is isomorphic to (R,+). Consider (x)=lnx. We know that ln is a 11
and onto function. From the properties of the logarithm we know that (ab) = ln ab = ln a +
ln b = (a) + (b), which means that ln preserves the multiplication. Notice that any other
logarithm, would work just as well, so there may be many different isomorphisms between
two isomorphic algebras.
Example 1.18. (Z3,) is not isomorphic to (ZOO,) (see Example 1.5). How does one prove
that something (in this case an isomorphism) does not exist? One way could be to check all
six bijections of Z3 onto ZOO and verify by hand that none preserves , but we can do better
than that. Suppose there exists an isomorphism  from Z3 onto ZOO. Notice that in Z3 12 =
21. So, if  is an isomorphism then (1)(2) = (12) = (21) = (2)(1). A quick
look at the table of the  operation in Example 1.5 tells us that in the ZOO algebra ab is
never equal to ba, so no matter what  assigns to 1 and to 2, (1)(2) is not equal to
(2)(1). Hence no such  exists.
Let us denote by  the relation of “being isomorphic to”, i.e. AB means that the
algebra A is isomorphic to B.
Theorem 1.1 The relation of “being isomorphic to” is an equivalence relation on any set of
binary algebras.
Proof. Any algebra is obviously isomorphic to itself, we simply use id for . Hence  is
reflexive.
Suppose now that (X,f1,f2, … ,fn) is isomorphic to (Y,g1,g2, … ,gn) and  is an
isomorphism. Then  is invertible and -1 is a bijection from Y onto X. Let us chose any two
elements a and b from Y. Since  is onto, there exist p and q in X such that a=(p) and
b=(q). Now:
-1(agib) = -1((p)gi(q))
{by the definition of p and q}
= -1((pfi q))
{because  is an isomorphism}
= pfi q
{from the definition of -1}
= -1(a)gi-1(b)
{by the definition of p and q, p=-1(a) and q=-1(b)}
Hence -1 is an isomorphism, so (Y,g1,g2, … ,gn) is isomorphic to (X,f1,f2, … ,fn) and that
means that  is a symmetric relation.
Finally, suppose that (X,f1,f2, … ,fn) is isomorphic to (Y,g1,g2, … ,gn) and (Y,g1,g2, …
,gn) is isomorphic to (Z,h1,h2, … ,hn), with  and  serving as isomorphisms. We will show
that ◦ is an isomorphism of (X,f1,f2, … ,fn) with (Z,h1,h2, … ,hn). Indeed, for any two a and
b from X and any subscript i we have
◦(afib) = ((afib))
{by the definition of function composition}
= ((a) gi (b))
{since  is an isomorphism}
= ((a)) hi ((b))
{since  is an isomorphism}
= ◦(a) hi ◦(b)
{by the definition of function composition}
Hence  is transitive. 