# Lesson 23: More Fun With Groups

In the previous lesson we did something rather drastic.  We considered what we were “really” doing when we added two numbers together—i.e., by saying that 5 plus 3 equals 8—and let that motivate the definition of a group.  However, the definition of a group is abstract, and need not apply at all to numbers and our desire to add them together.  A group is just a set with a special function—or way of combing elements in that set—along with a special element (with respect to that special function), and the property that for every element in the group, there is an element in the group that, when combined with it, gives the special element.

Now, it just so happens to be the case that the integers form a group, with the “combination rule” being addition, the special element being zero, and with the negative of any number being that element that, when combined with the original element, takes you back to the special element.  But we also saw, in the exercise for the previous lesson, that there are lots of other kinds of groups.  We saw that we could make a group out of rotations, or a group out of the numbers $\{0, 1, 2, 3, 4\}$, for example.  There are in fact lots and lots of other kinds of groups, and much of the power of the definition of a group comes from the fact that it shows us that when we’re adding numbers together, or multiplying fractions or real numbers (not including zero), or when we’re rotating a square, or when we’re doing the “clock arithmetic” briefly mentioned in the previous lesson and its exercise, we’re really doing the same thing: group theory.  Thus, whatever I can prove about a generic group, I can immediately prove about adding integers, multiplying fractions, rotating things, and whatever else constitutes a group.  This is what math is all about—finding unity in mathematical structures through abstraction.

In this lesson, I want to prove a couple of things about generic groups that might at first seem trivial, but which really illuminate the power of the definition of a group.  Namely, I will prove that “zero is unique” and that “there is only one negative”.  I’ll give those phrases more clear meaning when the time comes.  These proofs will help the reader get used to the sort of logic one uses in these arguments, and is a true introduction to the field of math known as “abstract algebra”.  Abstract algebra can, in many ways, be thought of as “reconsidering elementary school math”.  We’ll see that using both the set theory and the group theory that we’ve developed so far, we can give a firm and precise meaning to ideas that one likely took for granted until now.

Before doing so, however, I want to introduce some new notation.  This new notation will not only prove useful for us, but also bring out the abstraction that the notion of a group provides.  Suppose $G$ is a group, and let $g: G\times G\rightarrow G$ be its corresponding “special function”.  We know that when $G$ is $\mathbb{Z}$, we don’t often write $g(a,b)$.  Instead, we simply write $a+b$ to denote the same process.  The “$+$” sign is just shorthand for “take what comes before it and put it in the first slot of $g(\cdot, \cdot)$, and take what comes after it and put it in the second slot”.  But we don’t often think in those terms, and often times using $g(\cdot, \cdot)$ explicitly only confuses the matter.

Therefore, let me introduce the notation $a\cdot b$ to express $g(a,b)$.  Note that we can use this notation for any group.  In particular, when the group is $\mathbb{Z}$ with the obvious “group structure” of addition (here, “group structure” is what we call the operation $g:G\times G\rightarrow G$), then it just so happens that $a\cdot b=a+b$.  It also just so happens that $a\cdot b=b\cdot a$, but this is a very particular extra property that the integers have, and it is not included in the definition of a generic group.

Note that the definition of a generic group implies that $(a\cdot b) \cdot c=a\cdot (b\cdot c)$ for any $a,b,c\in G$.  This is the associativity property, and the above is nothing but a new notation for saying that it doesn’t matter if we apply the operation to the first two, and then to the third, or if we first apply it to the last two, and then to the first.  Remember, the operation takes two elements in $G$ (or one element in $G\times G$), and gives another element in $G$.  It is by no means necessary that a function must be associative, but since many interesting results come by assuming associativity, we include it in the definition of a group.  It might be the case that a set can come with a special function that is not associative, and that’d be fine, but we just wouldn’t call that set with that function a group.  We’ll see examples of non-associative functions soon.

(Note that, since a group’s special function is associative, we don’t have to worry about parentheses and we can unambiguously write $a\cdot b \cdot c$ instead of $(a\cdot b) \cdot c$ or $a\cdot (b \cdot c)$ since both of the latter expressions are the same.  As usual, this does not mean that we can write $a\cdot b \cdot c= a \cdot c \cdot b$, because it’s not necessarily the case that $b\cdot c=c\cdot b$, and therefore there is an ambiguity here about what elements we combine first.  Sometimes, when the special function is understood or implied, we can be even more sloppy and just write $ab$ instead of $a\cdot b$, with the former again meaning that $a$ goes in the “left slot” and $b$ goes in the right.  Then, we can unambiguously write $abc$, for example, but I’ll try not to get sloppy like that at least for a while, and when I do get sloppy I’ll try to let you know.)

We can refer to this “$\cdot$” thing as “the group multiplication”.  We must remember, though, that it might not have anything to do at all with multiplication of numbers.  This is just the abstract and general way of combining elements in some given group G.  If the group happens to be the fractions or the real numbers excluding zero, then the group multiplication is indeed regular multiplication.  But, if the group is the integers, then the group multiplication is actually addition!  And if the group is “rotations of a square”, as in the solution to the previous lesson’s exercise, then the group multiplication is actually “composition of rotations”.  This should hopefully bring out how abstract the group multiplication is.  Any group has such a multiplication rule, by its very definition, and we choose to denote it by $\cdot$, regardless of what it actually is.

The final thing we need to discuss about this notation is that the expression $a\cdot b$ must be viewed as a single element in the group $G$.  In other words, the expression $a\cdot b$ is a single expression that says “the result of multiplying $a$ by $b$”, where by “multiplying” I mean whatever the abstract group multiplication is in the given situation.  Thus, I can write an expression like $a\cdot b=c$ to say that the result of multiplying $a$ by $b$ is $c$, another element in $G$.

Now let us prove our first result—namely, that “zero is unique”.  What do I mean by this?  I mean that for any group $G$, the “special element” that it comes with is actually unique, i.e., there is only one.   Note the significance of this.  The fact that there is only one special element with the properties in the definition of a group is by no means obvious from the definition alone.  All we did was suppose that there was an element with those properties, but we never explicitly supposed that there was only one.  We can, however, prove that there is only one simply from the axioms of a group.  Let’s do this now.

Let $G$ be any group, and let $e$ and $f$ be elements in $G$ with the properties of a “special element”.  We’ll henceforth call these special elements “identities”, so that we can say that we’re letting $e$ and $f$ be identities in $G$.  We’ll show, only using the properties of a generic group, that $e=f$, so that they’re actually not different elements at all.

By the properties of an identity, it is the case that $e\cdot f=f$ because $e$ doesn’t change the thing it acts on, since it’s an identity.  Similarly, since $f$ is an identity, it must also be the case that $f$ fixes $e$, so that $e\cdot f=e$.  But the left hand sides of these equations are identical, so we have the following chain of equalities:

$e=e\cdot f=f$,

and therefore $e=f$ (by the transitive property of equality, I guess).  Thus, if we suppose that two elements are identities, then they must be equal, so that there is only one identity!

In the case of $\mathbb{Z}$, this means that zero is the only number that, when added to any other number, gives the other number back again.  Now, this may seem extremely obvious, but what’s important is that we’ve now developed a framework (with sets and groups) to actually prove that there is no other such element.  Thus, once we find one identity in a group, we can stop looking, because we know it’s the only one!

Let us now prove that “there is only one negative”.  By this I mean that for any element $a$ in some group $G$, there is only one element $b$ such that $a\cdot b=e$ and $b\cdot a=e$, where $e$ is the identity (and note that I can say “the” identity because we just proved it’s unique!).  We use very similar logic as the proof of the uniqueness of the identity.

Let $G$ be any group, and let $a$ be any element in that group.  Finally, let $b$ and $c$ have the property that they both “take $a$ back to the identity”.  We’ll show that $b=c$, thus proving that there is only one such element.

We know that $a\cdot b=e$ and that $a\cdot c=e$ by definition of $b$ and $c$ and the fact that we now know that the identity unique.  Thus, we can set these two equal to each other:

$a \cdot b=a \cdot c$

Remember, this is a statement of equality between two individual elements in $G$.  Thus, if we multiply (“abstract group multiply”) both sides by the same element, the resulting expressions will be equal as well because we’re simply multiplying a single element in $G$, written two different ways, by the same element.  Thus we have, in particular, that

$c\cdot (a\cdot b)=c \cdot (a \cdot c)$

where all we did was “multiply by $c$” on the left of both sides of the equality (which again means “apply the special function to $c$ on the left and the element $a\cdot b=a\cdot c$ on the right).  Since our combination rule is associative, we can move these parentheses back, as follows (since the order of combination doesn’t matter):

$(c\cdot a)\cdot b=(c \cdot a) \cdot c$.

But by the definition of $c$, we know that $c\cdot a=e$, and thus we have that

$e\cdot b=e\cdot c$,

and since $e$ is the identity, it fixes these elements (i.e., $e\cdot b=b$ and $e\cdot c=c$), and we have that $b=c$, which is what we wanted to prove.  Thus, there is only one such element that behaves this way in relation to $a$, and we call it the inverse of $a$ and usually denote it by $a^{-1}$.  Note, though, that $a^{-1}$ is nothing but a single expression for some element in $G$, and we use this notation simply to remind us that it has this special property with respect to $a$.

We’ve thus shown that once we’ve found one inverse of an element, we can stop looking for another one, because there will only be one.  I.e., once we know that $(-4)+4=0$, we can stop looking for another inverse of $4$.  Again, this might seem completely obvious, but “obviousness” is hardly a mathematical proof.  Certain mathematical ideas are “obvious” simply because of various experiences that we’ve had as humans that make them seem obvious.  We must be able to find a formalism that proves these statements rigorously, and group theory does that.  It also is one of many steps towards seeing what’s “really” going on even when we make the most simple mathematical moves, like adding or subtracting.

But don’t worry—group theory does much, much more than simply prove obvious statements and reformulate basic algebra.  After all, the true power of group theory does not come from its ability to revamp 3rd grade math.  Instead, its true power comes from its abstraction and the insight it provides into math that is far beyond that of the 3rd grade.  It is an immensely rich field, and we’ll be exploring lots of different aspects of it for a while.  Thus, if you have yet to really see the power or beauty in this definition, don’t fret, there will be much cooler applications of group theory to come.  I’ll leave one small result as an exercise (with solution, of course), and in the next lesson we’ll start to explore the sub-structure of groups.

Exercise:

1)  Let $G$ be a group and let $a$ be any element in $G$.  Prove that the inverse of $a^{-1}$ is $a$.  I.e., show that “the inverse of the inverse is what you started with”.  (Hint: this is completely trivial from the definition of a group.  I.e., if we know that $a^{-1}$ is the inverse of $a$, then we simply take the properties that it has with respect to $a$ and note that they imply that $a$ also immediately has those properties with respect to $a^{-1}$.  This is mainly an exercise in getting the flow of logic correct and precise.)

Solution to Exercise

On to Interactive Interlude

On to Lesson 24

Back to Lesson 22

### 3 Responses to Lesson 23: More Fun With Groups

1. Ahmed says:

The abstract algebra, Is it the new approach of Mathematics?
Is there an abstract calculus?

• It’s not so much about the NEW approach to math, but rather just THE approach to math. Abstract algebra just puts on a more rigorous footing all that we’ve done when we’ve done algebra before. I.e., it allows us to more clearly answer the question “What are we REALLY doing when we do algebra?”, and in exploring that question we not only partially answer it, but we find a much more general (i.e., abstract), form of algebra. Accordingly, there is an “abstract calculus” in a sense, and it is called “real analysis”. However, it isn’t a NEW calculus, but rather an answer to the question of what we’re “REALLY” doing when we do calculus. Thus, calculus is a small subset of analysis, but the abstract foundations of analysis allows us to do much more than calculus alone lets us do, in the same way that what we thought of as “algebra” really just “sits inside of” abstract algebra as a small subset.

• Ahmed says:

Thanks a lot. It is very helpful.