This is a version of an elementary lecture I gave recently. At the end some questions raised or implied but not answered in the lecture are posed as exercises; these are cited along the way as Q1, Q2, etc. Some of them are less gentle than the lecture.

If symbols are not legible, refreshing the page usually restores them.

Part 1.  Groups of real numbers

1.0  Real numbers

Consider the collection, or set, of all the numbers—what we usually mean by “numbers,” the so-called real numbers, positive, negative, and zero, comprising the whole numbers (integers), the fractions, and other numbers such as $\sqrt 2$ and $\pi.$ These are the numbers which we often think of as associated to all the points of the number line. Let us call this collection $R.$ Instead of calling its members numbers, we may call them elements; in general, the term element may be used of any member of a set of things.

1.1  Groups of real numbers

$R$ comes equipped with two principal operations, or laws of composition, namely addition and multiplication. These have certain properties in common.

1. They are associative, which means that how elements are grouped does not affect the result of performing the operations. That is, in the case of addition, for any elements $a,b,c,$ $(a+b)+c=a+(b+c);$ hence the parentheses are unnecessary, and we write simply $a+b+c$ for the sum of three numbers. Likewise, for multiplication, $(ab)c = a(bc),$ so we write $abc.$

1a. They are commutative, which means that the order of elements does not affect the result of performing the operations. Thus for any $a$ and $b,$ $a+b=b+a$ and $ab=ba.$

2. Each operation has a two-sided identity element, namely $0$ for addition and $1$ for multiplication, with the property that combining it by the operation with any element leaves that element unchanged (identical to what it was). That is, for any $a, 0+a=a$ and $1\times a=a.$ Two-sided means that combining with the identity gives the same result regardless of order: $a+0=a$ and $a\times 1=a.$ Of course, this follows from no. 1a above; it is included here because later we will consider operations that may not be commutative.

3. For each operation, every element $a$ has a two-sided inverse, an element whose combination with $a$ by the operation, in either order, is the identity; with one exception, to be dealt with in a moment. For addition, the inverse of $a$ is $-a:$ $(-a)+a=0.$ For multiplication, the inverse of $a$ is $1/a:$ $(1/a)\times a=1.$ The latter inverse is also written $a^{-1},$ and that is the notation we will be using: $a^{-1}a=1.$

In the case of multiplication, there is one element which has no inverse, namely $0.$ In comparison to addition, therefore, multiplication can be regarded as defective: it fails to possess property 3 completely. We may undertake to remove the defect by confining multiplication to a part of $R$ which does not contain the offending element. Since from our present (limited) point of view $0$ and the negative numbers serve merely to provide an identity and inverses for addition, if our concern is with multiplication alone we can cast them out, leaving only the positive real numbers, $R^{pos}.$ In this set every element has a multiplicative inverse, and the other properties, 1, 1a, and 2, still hold as well. Alternatively, we can be thriftier, removing from $R$ only the element $0$ that caused the problem. By so doing we obtain the non-zero real numbers $R^{\neq 0}.$ In this set too multiplication has all of the properties 1, 1a, 2, and 3.

Because they have properties 1, 2, and 3, each of the following three sets-with-operations is called a group (Q1). And because they also have property 1a, each is an abelian group. (The adjective “abelian” means only that the operation is commutative.)

I. $R,$ with the operation of addition

II. $R^{pos},$ with the operation of multiplication

III. $R^{\neq 0},$ with the operation of multiplication

Group I can be called an additive group (that is, a group whose operation is written as addition), groups II and III multiplicative groups. We speak of $R,$ $R^{pos},$ and $R^{\neq 0}$ as the sets underlying the groups. Often a group is referred to by its underlying set, when the operation is understood; thus in our context “the group $R$” means the additive group of real numbers, group I.

1.2  Distinguishing groups

If we view our three sets of real numbers purely as the groups I, II, III, without regard to any other properties they have, we may ask: as groups, are they distinguishable? Is the behavior of their elements under the group operations alone sufficient to tell them apart? Are they somehow groups of different types?

In order to investigate this question, let us begin by applying each group operation in the simplest way, to a single element.

I.  Given an arbitrary element $a$ of the additive group I, we first add it to itself, obtaining $a+a= 2a,$ then add it to that result, obtaining $3a,$ and so on, so that we produce the sequence of numbers $na$ for all the positive integers $n.$ What can be said about this sequence? There are two cases. If $a$ is the identity element $0,$ every one of the numbers $na$ is $0.$ If $a$ is not $0$, none of these numbers is $0.$ For if $a$ is positive, $2a$ is greater than $a,3a$ is greater than $2a,$ and so forth; while if $a$ is negative, $2a$ is more negative than $a,3a$ is more negative than $2a,$ and so forth.

II.  Given an arbitrary element $a$ of the multiplicative group II, we proceed in the analogous way, multiplying it by itself to get $a\times a = a^{2},$ then $a^3,$ and so on, thus obtaining $a^n$ for every positive integer n. As before, there are two cases. If $a$ is the identity element $1,$ every one of the numbers $a^n$ is $1.$ If $a$ is not $1,$ none of these numbers is $1.$ For if $a$ is greater than $1,a^2$ is greater than $a,a^3$ is greater than $a^2,$ and so forth; while if $a$ is less than $1,a^2$ is less than $a$ (for example, if $a=1/3, a^2=1/9$), $a^3$ is less than $a^2,$ and so forth.

III.  Group III is multiplicative too, so we again construct the sequence $a, a^2, a^3,\cdots;$ that is, $a^n$ for every $n.$ As before, if $a$ is the identity element $1,$ every one of the numbers $a^n$ is $1.$ If $a$ is not $1,$ we might at first expect that as in the preceding cases, none of these numbers could be $1;$ and this is almost true, but there is one exception: when $a=-1, a^2=1.$

We say that the element $-1$ of group III is of order $2,$ because $2$ is the least exponent of $-1$ which makes it equal to the identity; that is, $-1$ to the second power, but not to the first, equals $1.$ More generally, for any positive integer $n,$ an element $a$ of any multiplicative group is of order $n$ if $a^n,$ but no lower power of $a,$ equals the identity. If this is the case for some $n,a$ is said to be of finite order. Correspondingly, for an additive group, an element $a$  is of order $n$ if $na,$ but no lower multiple of $a,$ equals the identity. Of course in any group the identity itself is of order $1.$

We have shown that in groups I and II there are no elements of finite order apart from the identity, whereas in group III there is an element of order $2.$ Thus we have a partial answer to the question asked above: on the basis of its character as a group alone, group III can be distinguished from the other two. It contains an element whose behavior under the group operation is unlike that of any element in the other groups.

1.3  Isomorphism of groups

Whether I and II can be distinguished from one another is as yet undecided. This question is precisely addressed by the concept of an isomorphism between two groups, which means (i) a one-to-one correspondence between the sets of elements, that (ii) respects or preserves the group operations. (This will be clarified presently.) We are asking whether I and II are isomorphic—whether there exists an isomorphism between them. If they are isomorphic, they cannot be distinguished by any test which relies solely on the behavior of their elements under the group operations.

An isomorphism between groups I and II, then, would be in the first place a one-to-one correspondence between their underlying sets; that is, a correspondence in which each element in one set is matched with exactly one element in the other:

$R\longleftrightarrow R^{pos}$              (1)

(Q2). Secondly, suppose that under this correspondence an element $a$ of $R$ corresponds to the element $a^{\prime}$ of $R^{pos},$ and an element $b$ to $b^{\prime}:$

$a\longleftrightarrow a^{\prime}$

$b\longleftrightarrow b^{\prime}.$

Then for the correspondence to define an isomorphism, we must also have

$a+b\longleftrightarrow a^{\prime}\times b^{\prime};$

and this will be the case for every $a$ and $b$ in $R$ and their counterparts in $R^{pos}.$ The result of combining two elements in one set corresponds to the result of combining the corresponding two elements in the other set. One group operation mimics the other.

It is an easy consequence of the definition that under an isomorphism identity elements and inverses correspond (Q3):

$0\longleftrightarrow 1$

$-a\longleftrightarrow a^{\prime -1}.$

1.4  Non-isomorphic groups

Before undertaking to discover whether $R$ and $R^{pos}$ are isomorphic, let us show that $R^{\neq 0}$ and $R^{pos}$ are not isomorphic, by formulating the argument of §1.2 in terms of isomorphism. The proof will be by contradiction.

Suppose that there is an isomorphism

$R^{\neq 0}\longleftrightarrow R^{pos}.$

Then the identities correspond,

$1\longleftrightarrow 1.$

(In this case, unlike the one in §1.3, both groups are multiplicative, so both identities are $1.$) The element in $R^{\neq 0}$ that we used before to distinguish it from the other two groups corresponds to some element, call it $x:$

$-1\longleftrightarrow x.$

Here $x \neq 1,$ because in the one-to-one correspondence $-1$ must correspond to a different element from the one $1$ corresponds to. But since the operations correspond, we also have

$-1\times -1\longleftrightarrow x\times x.$

Inasmuch as $-1\times -1=1,$ and $1$ corresponds to $1,$ this implies that $x^2=1.$ But $x$ is a positive number, and the only positive number whose square is $1$ is $1$ itself. Therefore $x=1,$ which is a contradiction.

The same argument can be used to show that $R^{\neq 0}$ is not isomorphic to $R$ either (Q4).

1.5  Isomorphic groups: exponential function and logarithm

Now we consider the possibility of the isomorphism proposed in §1.3,

$R\longleftrightarrow R^{pos}.$             (1)

For this purpose we adopt functional notation, denoting the proposed isomorphism in the direction from left to right by the letter $f,$ so that corresponding elements are written as follows:

$a\longleftrightarrow f(a)$

$b\longleftrightarrow f(b).$

Then on the one hand

$a+b\longleftrightarrow f(a+b),$

and on the other

$a+b\longleftrightarrow f(a) \times f(b).$

Hence the equation

$f(a+b)=f(a)f(b)$             (2)

holds for every $a$ and $b$ in $R.$ It is a trivial matter to extend equation (2) to more than two summands:

$f(a+b+c)=f((a+b)+c)=f(a+b)f(c)=f(a)f(b)f(c),$

and in general

$f(a_{1}+a_{2}+\cdots +a_{n})= f(a_{1})f(a_{2})\cdots f(a_{n}).$                     (3)

This equation limits the forms $f$ can take. For consider first $f(m),$ where $m$ is any positive integer. Writing $m$ as the sum of $m$ ones and applying (3), we have

$f(m) =f(1+1+\cdots+1)=f(1)f(1)\cdots f(1),$

with $m$ factors. If for the sake of economy we introduce the notation $c=f(1),$ the preceding equality becomes

$f(m)=c^{m}.$

The value of $f$ on a positive integer $m$ is simply $c$ raised to the power $m.$ Note that $c\neq 1,$ because $f(1)\neq f(0)=1.$

Next consider the value of $f$ on an arbitrary positive fraction $\frac{m}{n}$ , where $m$ and $n$ are positive integers. Since

$n\frac{m}{n}=m,$

we have

$f(m)=f(n\frac{m}{n})= f(\frac{m}{n}+\frac{m}{n}+\cdots +\frac{m}{n})=f(\frac{m}{n})^ {n},$

or

$f(\frac{m}{n})^ {n}=c^{m}.$

Taking $n$th roots of both sides, we obtain

$f(\frac{m}{n})=c^{\frac{m}{n}}.$

This formula, now proven for all positive fractions, obviously holds for $0$ as well, and can easily be extended to negative fractions. Thus for all rational numbers—numbers which can be expressed as fractions—the isomorphism has the form

$f(x)=c^{x},$                   (4)

where $c\neq 1$ is a fixed but undetermined positive number (Q5). Since the rational numbers are everywhere thickly distributed among the real numbers—we say that they are dense in $R,$ meaning that in every interval of real numbers, no matter how small, rational numbers are to be found—this strongly suggests that formula (4), as applied to all real numbers, should be tested to see whether it provides an isomorphism as in (1). In fact it can be shown that it does, for any value of $c;$ and further, that there is a best (i.e., most convenient) choice for $c,$ namely a number called $e,$ whose value is approximately $2.718.$ (The demonstration of these facts can be found in books on calculus and analysis.). Thus our groups I and II are indeed isomorphic, and the standard isomorphism in the direction from $R$ to $R^{pos}$ is given by the exponential function $e^{x},$ which is also written $\exp x:$

$R\longrightarrow R^{pos}$

$x\longrightarrow\exp x.$

In the other direction the isomorphism is called the logarithm, $\log y:$

$R\longleftarrow R^{pos}$

$\log y\longleftarrow y.$

Hence we have the familiar law of exponents

$\exp (a+b)=\exp a\ \exp b,$

and the corresponding property of the logarithm,

$\log (ab)=\log a +\log b.$

These relations, which connect the two fundamental operations on real numbers, are a primary reason why the functions $\exp$ and $\log$ occur everywhere in mathematics and its applications.

The two functions have many other good properties besides defining the group isomorphism. In particular, they preserve the order of the numbers (“order” in the usual sense of arrangement by increasing size): if $a and likewise for $\log.$ But seeing that the sets $R$ and $R^{pos}$ are manifestly not the same in all respects, it cannot be that the functions establish perfect identity between them. Nor do they, since they fail to preserve the difference between numbers. For example,

$1-0=1$ but $\exp 1 -\exp 0 =e-1\approx 1.718.$

That difference cannot be preserved by any order-preserving one-to-one correspondence between $R$ and $R^{pos}$ is evident from the number line, where difference measures the distance between points (Q6).

Part 2.  The smallest groups

2.0  Finite groups

Let us now turn from the familiar infinite to the unfamiliar finite. A finite group is one with finitely many elements. In this second part all groups will be finite, and we will be using multiplicative notation for them. The identity element of a group will be denoted by the letter $e,$ a standard notation which has nothing to do with the number $e$ that defines the exponential function. Thus for any element $a,$ $ea=ae=a$ and $a^{-1}a= aa^{-1}=e.$

A group will be indicated by listing its elements within curly brackets, thus: $\{e,a,b,c\}.$ The number of elements in a group is called its order; as will soon become clear, this use of the term is closely related to the sense introduced in §1.2.

In a statement such as “There are exactly two groups of order $4$” the qualification “up to isomorphism” is to be understood. Strictly speaking, such a statement refers not to specific individual groups, but to isomorphism classes of them. Given two non-isomorphic groups $F$ and $G$ of order $4,$ to say that these are the only groups of that order means that all groups of order $4$ are isomorphic either to $F$ or to $G.$

To determine a group of given order up to isomorphism, it suffices to set up the multiplication table for the group—to establish how all of the elements combine with one another under the law of composition. For the multiplication table completely determines that law, which any isomorphic group equally obeys. The examples that follow will illustrate this.

2.1  Groups of orders 1 and 2

Order 1

There is only one group of order $1,$ namely $\{e\},$ with the rule $e\times e=e.$

Order 2

A group of order $2$ has the form $\{e, a\},$ where $a\neq e.$ We know what $ee,ea,$ and $ae$ are; what is $aa$? There are two possibilities: $aa=a$ or $aa=e.$ But if $aa=a,$

$a^{-1}aa=a^{-1}a,$

whence $ea=e,$ or $a=e,$ which is not the case. Therefore $a^{2}=e,$ and the multiplication table is fully determined by the fact of there being just two elements; in other words, there is only one group of order $2.$ It is obviously abelian.

Small though it is, the group of order $2$ is very important. One reason is that it can be regarded as describing an on-off switch—one that has two states, off and on. Given such a switch—a push-button, say—there are two actions that can be taken. One can not push the button, or one can push it. (The former of these would ordinarily be regarded as inaction rather than action; we regard it as null action.) Not pushing leaves the state of the switch, whether off or on, unchanged; pushing changes the state. Each action can be thought of as a transformation of the state of the switch, and two transformations can be combined by successive performance to yield a single transformation which has the same effect. The order of performance does not matter. Thus (i) not pushing followed (or preceded) by pushing is the same as pushing; (ii) pushing followed by pushing is the same as not pushing; etc. (We are not concerned with any effect that pushing the switch might have on something it controls, only with the switch itself.)

Let us represent the identity element $e$ of our group by the action of not pushing (which leaves the state of the switch identical to what it was), and the element $a$ by pushing; further, let us represent multiplication in our group by successive performance of actions. Under this interpretation, the group is exhibited as a transformation group, the two-element set of transformations of the state of the switch, transformations being combined by successive performance. The multiplication rule $ea=a$ corresponds to (i) just above, the rule $a^{2}=e$ to (ii), and so forth.

Another application of this group is to a pair of things, such as the letters $p$ and $q,$ which can stand in either of two orders: $p ~~~ q$ or $q ~~~ p.$ (Here we are again using the word “order” in an ordinary sense.) To permute the letters means to change their order, and there are two possible permutations: one can not change the order, or one can change it. As with the on-off switch, permutations can be combined by successive performance. If we represent $e$ by not changing the order, $a$ by changing it, and multiplication by successive performance, we have a representation of our group as a permutation group on two letters. Lest this seem unimportant, it may be mentioned that the fact that a certain permutation group, a so-called Galois group, has order $2$ rather than $1$ is equivalent to the incommensurability of the diagonal and side of a square, the discovery of which so excited the ancient Greeks.

A numerical example of the group of order $2$ is the multiplicative group $\{1,-1\}.$ With the aid of this group we can express the group $R^{\neq 0}$ in terms of the group $R^{pos},$ to which it was shown in §1.4 not to be isomorphic. To do this, we define a new group, the direct product of $\{1,-1\}$ and $R^{pos},$ written

$\{1,-1\}\times R^{pos},$

as follows. The elements of the direct product are the ordered pairs (pairs written in a definite order) $(s,a),$ where $s$ is an element of $\{1,-1\}$ and $a$ is an element of $R^{pos},$ and multiplication of pairs is defined by the rule

$(s,a)\times (t,b)=(st,ab).$

It is easy to show that the direct product is a group, and that an isomorphism

$\{1,-1\}\times R^{pos}\longrightarrow R^{\neq 0}$                    (5)

is defined by

$(s,a)\longrightarrow sa$

(Q7).

2.2  Groups of orders 3, 4, and 5

Orders 3 and 5

A group of order $3$ has the form $\{e,a,b\}.$  Here $a^{2}$ must equal $b,$ because just as in the order $2$ case $a^{2}\neq a,$ while if $a^{2}=e,$ which of the three elements will $ab$ be? It cannot equal $e,$ because that would imply

$b=eb=aab=ae=a.$

Nor can $ab$ equal $a,$ for in that case we would have

$b=eb=a^{-1}ab=a^{-1}a=e.$

Similarly, $ab$ cannot equal $b.$ Thus the only possible value of $a^{2}$ is $b,$ so there is only one group of order $3,$ and it has the form $\{e,a,a^{2}\}.$ It is readily seen that $a^{3}$ cannot equal $a$ or $a^{2},$ hence $a^{3}=e.$

The group of order $3$ is called a cyclic group, a designation which can be understood as signifying that the successive powers of one element cycle through all the elements of the group: $a^{1}=a,a^{2}=a^{2},a^{3}=e.$ A cyclic group, then, is one that contains an element whose order (in the sense of §1.2) is equal to the order of the group.

It is obvious that the groups of orders $1$ and $2$ are likewise cyclic. The numbers $2$ and $3$ are prime, and it is a fact that every group of prime order is cyclic. This follows from a general proposition, that in any finite group, the order of every element divides the order of the group (Q11). (Here divides means “is a divisor of,” “is a factor of.”) Given this, we conclude that as with order $3,$ there is only one group of order $5,$ namely a cyclic group of the form $\{e,a,a^{2},a^{3},a^{4}\}.$

There are cyclic groups of all orders, since given any positive integer $n,$ we can simply define a group of $n$ elements $\{e,a,a^{2},a^{3},\cdots, a^{n-1}\}$ by modifying the usual law of exponents for multiplication with the additional rule $a^{n}=e.$ For each $n$ there is only one cyclic group of order $n,$ and it is abelian. It can be represented as a group of transformations of the state of a dial with $n$ positions, multiplication by $a$ corresponding to turning the dial one step in a fixed direction, say counterclockwise.

Order 4

Besides the cyclic group there is one other group of order $4,$ namely the abelian group $\{e,a,b,c\}$ with $a^{2}= b^{2}= c^{2}=e$ and $ab=c,bc=a,ca=b.$ It can be represented as a group of transformations of the state of two on-off switches. For given pushbuttons $S_{1}$ and $S_{2}$ there are four actions that can be taken: (i) push neither, (ii) push $S_{1},$ (iii) push $S_{2},$ (iv) push both. Representing $e$ by (i), $a$ by (ii), $b$ by (iii), $c$ by (iv), and multiplication by successive performance, as in §2.1 we obtain the desired representation of the group.

It is not difficult to show that there are no other groups of order $4$ (Q8).

Within the non-cyclic group we observe that there are three copies of the group of order $2,$ namely $\{e,a\},$ $\{e,b\},$ and $\{e,c\}.$ Each of these is a subgroup of the larger group, which means a part of a group which itself forms a group, under the same law of composition. A part of group forms a subgroup if it contains the identity element, the product of any two elements in it, and the inverse of every element in it.

2.3  Groups of order 6

As with order $4$, there is only one group of order $6$ apart from the cyclic group. The demonstration of uniqueness is more difficult in this case; it can be carried out directly (Q8), or obtained with the aid of certain general results not provided here. As for existence, rather than beginning with the multiplication table of the group, let us derive it as a group of transformations of a geometrical object.

The object is an equilateral triangle, and the transformations to be considered are its symmetries. A symmetry of the triangle, as of any figure, is the alteration effected by a rigid motion—a moving of the triangle without changing its shape or size (thus with no bending, breaking, stretching, or shrinking)—which returns it to the place it was in before; that is, which brings it back into coincidence with itself as it was before the motion. As the figure below shows, there are six of these symmetries, including the identity symmetry $e.$[1]

We have denoted by $a$ rotation counterclockwise through $120\textdegree$—more precisely, the effect of that rotation; then $a^{2}$ is rotation through $240\textdegree.$ (The former of these has the same effect as a clockwise rotation through $240\textdegree,$ the latter as a clockwise rotation through $120\textdegree.$) There are also three reflections in the axes of symmetry indicated by the dashed lines (i.e., rotations out of the plane about those axes). Of these we have arbitrarily chosen one to be called $b.$ Clearly $a$ is of order $3,$ and $b$ of order $2.$ The reflection below $b$ is given by the combination $ab$—first perform $a,$ then $b$—and also by $ba^{2};$ and the third reflection is given by $ba,$ and also by $a^{2}b.$ The relations $a^{3}=e,$ $b^{2}=e,$ $ab=ba^{2},$ and $ba=a^{2}b$ completely determine multiplication in the symmetry group $\{e,a,a^{2},b,ab,ba\}$ represented by the symmetries of the triangle. Known as the dihedral group of order $6,$ it is generated by the two elements $a$ and $b,$ called generators: all elements are expressible in terms of them. The same is true for other pairs of generators, for example the elements written $ab$ and $ba$ in our notation (Q9).

Here for the first time we encounter non-commutativity: $ab\neq ba.$ The dihedral group of order $6$ is the smallest non-abelian group. It contains one subgroup of order $3,$ namely the rotations (including the identity) $\{e,a,a^{2}\},$ and three subgroups of order $2,$ one for each reflection: $\{e,b\},$ $\{e,ab \},$ $\{e,ba\}.$ That these orders divide $6$ is a consequence of a general proposition, that in any finite group, the order of a subgroup divides the order of the group (Q10).

Why is the number of symmetries of the triangle six? Because a symmetry is a matter of assigning three vertices to three vertices. For example, $a$ assigns the vertex originally labeled $D$ to the vertex originally labeled $O,$ etc. A first vertex ($D,$ say) may be assigned to any of three vertices; once that is done, a second vertex ($O,$ say) may be assigned to either of the remaining two vertices; the assignment of the third is then determined. There are three choices for the first assignment, and for each of these, two for the second; hence six in all.

Thinking of the symmetries in this way suggests a different, non-geometrical, interpretation of the dihedral group, as the permutation group (§2.1) on the three letters $D ~~~ O ~~~ G.$ Assigning each letter to the one below it, we will have

$e$
$D ~~~ O ~~~ G$
$D ~~~ O ~~~ G$

$a$
$D ~~~ O ~~~ G$
$O ~~~ G ~~~ D$

$b$
$D ~~~ O ~~~ G$
$O ~~~ D ~~~ G$

and so forth.

We can say that the same abstract group of order $6$ acts on the triangle and on the letters, much as the abstract number 6 counts six eggs and six apples. What exactly “abstract” may mean is harder to say.

Concluding remarks

There is a great variety of groups, including transformation groups, geometrical and other, both finite and infinite. As an example of an infinite transformation group we note that the additive group $R$ introduced in §1.1 acts on its own underlying set of real numbers, or equivalently on the number line, by translation: an element $a$ of $R$ is interpreted as the action “add $a$ to each number,” or “shift rightward by the distance $a.$” A similar action describes the classical transformation of velocities in kinematics; it was later replaced by a more complex relativistic one. Symmetry groups apply not only to linear and planar figures, but also to objects in higher dimensions, and are invoked in physical science to describe everything from the structure of crystals to the society of fundamental particles (Q12).

Since groups of prime order seem to be the simplest ones (§2.2), we might wonder whether they play a role among all the groups analogous to that of the prime numbers among the positive integers—that is, whether any group can somehow be constructed from, and decomposed into, groups of prime order, as the number $60$ is expressed as the product of primes $2\times 2\times 3\times 5$ (Q13). That is not the case, but there is a larger class of so-called simple groups in terms of which finite groups, at least, can be decomposed. A simple group is one which has no non-trivial factors, in a certain sense of that word not unrelated to the usual one (Q14); thus its definition, which arises from work of Galois in 1832, is formally similar to that of a prime number. But to identify and classify even the finite simple groups is a profound problem, first explicitly posed in 1892, and completely solved only in 2008, after decades of labor by a host of mathematicians.[2] The solution is a remarkable intellectual achievement.

Questions

Q1. Suppose it is given that a set with an operation satisfies condition 1 of §1.1, but only the following weaker forms of conditions 2 and 3: there is a left identity element $e,$ i.e., one such that $ea=a$ for every $a;$ and every element $a$ has a left inverse, i.e., an element $a^{-1}$ such that $a^{-1}a=e.$ Are conditions 2 and 3 (in their full strength) then satisfied as well?

Q2. (a) Show that the set of positive integers can be put in one-to-one correspondence with a subset, or part, of itself. (b) Show that the same is true of every infinite set.[3] Thus a correspondence like (1) in §1.3, which matches the set $R$ with a subset of itself, is not a priori impossible.

Q3. Show that under an isomorphism identity elements and inverses correspond.

Q4. Rewrite the argument of §1.4, replacing $R^{pos}$ by $R.$

Q5. Establish formula (4) of §1.5 for negative fractions.

Q6. Show that difference cannot be preserved by any order-preserving one-to-one correspondence between $R$ and $R^{pos}$ (a) geometrically, using the number line, and (b) arithmetically.

Q7. (a) Verify isomorphism (5) in §2.1, defining it in the other direction. (b) Interpret the isomorphism in terms of the number line.

Q8. Show that: (a) there are only two groups of order $4,$ both of them abelian; (b) there are only two groups of order $6.$

Q9. Show that the dihedral group of order $6$ is generated by any two reflections.

Q10. Let $H$ be a subgroup of a finite group $G.$ For any element $g$ of $G$ define the left coset $gH$ to be the set of all the products $gh,$ where $h$ runs through all the elements of $H.$ Show that: (a) all left cosets have the same number of elements; (b) two left cosets, if not identical, have no elements in common; (c) the order of $H$ divides the order of $G.$

Q11. Show that: (a) the powers of any element of a group form a cyclic subgroup; (b) the order of any element of a finite group divides the order of the group. (c) What subgroups does a cyclic group have?

Q12. (a) Determine the symmetry group of the square. (b) Describe the symmetry group of the circle, specifying a set of generators.

Q13. (a) Show that the non-cyclic group of order $4$ is the direct product (§2.1) of two groups of order $2.$ (b) Which group of order $6$ is the direct product of the group of order $2$ and the group of order $3$? (c) Generalize to the direct product of groups of unequal prime orders. (d) Generalize further to groups whose orders are relatively prime, i.e., have no factors besides $1$ in common.

Q14. Let $H$ be a subgroup of a finite group $G$ whose order is $>1.$ Define the right cosets $Hg$ analogously to the left cosets (Q10). $H$ is a normal subgroup if for every $g$ in $G,$ $gH=Hg.$ The group $G$ is simple if it has no normal subgroups other than $\{e\}$ and $G$ itself. (a) If $H$ is normal, show that the cosets of $H$ form a group under the operation $(aH)(bH)=abH.$ This group is called the factor group $G/H.$ What is its order? (b) Which of the groups of orders from $2$ to $6$ are simple? (c) For the ones that are not, determine their normal subgroups and the corresponding factor groups.

[1] Thanks to Joshua Comenetz for producing the figure.

[2] See http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups , especially the timeline, and the following article cited there: Solomon, Ronald (2001), “A brief history of the classification of the finite simple groups”, American Mathematical Society. Bulletin. New Series 38 (3): 315–352, doi:10.1090/S0273-0979-01-00909-0 .

[3] One might not realize that the argument depends on a weak form of the set-theoretic axiom of choice.