Proving Certain Polynomials Form a Group

Abstract algebra can be a huge leap for many students, who may know algebra well, but are not used to abstraction – generalizing the concept of numbers so we can invent new kinds of “numbers” and “operations” and comparing their properties. Here we will look at a question from a student beginning the study of “groups”, working with her through some of the challenges of the new subject.

The problem: Prove this is a group

Here is the initial question, from early April:

I need to solve this problem:

I know the 4 things that need to be proven, and associative has been completed as sidework by the professor. I just do not know how to even start the problem.

A group is “a set equipped with an operation that combines any two elements to form a third element while being associative as well as having an identity element and inverse elements.” Hidden here is a fourth requirement, that the operation must be closed (that “third element” must be a member of the same set). The professor took care of what is sometimes the hardest part, proving associativity (we’ll take a look at that at the end), leaving three for Jennifer to prove.

The set under discussion, called \(\hat{P}_2\), is a particular set of polynomials, namely those that can be written in the form \(ax^2 + b\) with real coefficients (second-degree real polynomials lacking a linear term). The big challenge for the newcomer is that the operation called “*” is not ordinary multiplication of polynomials, but something peculiar, made up for this problem. (From experience, it happens that I immediately recognized this operation as something familiar in disguise. We’ll take a look at that, too.) But for now, we need to help Jennifer!

Defining a group

Doctor Fenton answered her, just checking what she knows by carefully stating what has to be done:

Hi Jennifer,

The three properties you need to prove are (1) closure; (2) existence of an identity; and (3) existence of inverse elements.

The statement gives you the definition of the underlying set P2ˆ(x): it consists of polynomials of the form ax2+b, where a and b are real numbers and a ≠ 0, i.e. real polynomials whose degree is exactly 2.  It also defines the binary operation p(x)*q(x)  (Note that this operation is NOT the usual polynomial multiplication!)

(1) Closure: Is the product p(x)*q(x) an element of P2ˆ(x)?  That is, does it satisfy the conditions to be in P2ˆ(x)?

(2)  Is there an identity element? That is, is there a polynomial e(x) such that for every p(x) in P2ˆ(x), e(x)*p(x) = p(x)?   (If there is such an element e(x) = cx + d, what properties must the coefficients c and d satisfy?)

(3)  Does every element in P2ˆ(x) have an inverse? Given p(x) = ax2 + b, is there an element q(x) = cx + d such that p(x)*q(x) = e(x)?

Can you answer any of these questions?

Jennifer replied,

Thank you. That is a better explanation of what they’re looking for. I understand the three things I’m supposed to do; I just don’t understand how to start doing them. I don’t understand how to prove that there’s an inverse. I don’t understand how to show that there is an identity. I’m actually completely lost.

This is not very surprising! She has presumably seen examples of how to prove something is a group, but every example is so different, it can be hard to see how to apply examples to a new problem.

Looking for an identity element

Where shall we start? It might have been most natural to start by demonstrating what closure means in this particular case, but Doctor Fenton chose to start with the the identity and the inverse, which Jennifer had explicitly asked about:

You have to know what the identity element is before you can find an inverse.  Suppose p(x) = ax2+b.

An identity element will be an element e(x) = cx2 + d, and it must satisfy p(x)*e(x) = p(x).

Compute the product on the left side.  It gives you a polynomial will coefficients given by the product formula in the problem.  That product must be the same as p(x).  Can you see values of c and d which will make that true?  (The coefficient of x2 in the product must be the same coefficient as x2 in p(x).  What equation does that give you?) Similarly, the constant term in the product must be the same as the constant term in p(x).  You need to find values of c and d which will make these equations true for all values of a and b.

So we are trying to find a polynomial such that “multiplying” any polynomial in the set by it will leave it unchanged. He is suggesting writing a pair of equations and solving for the needed coefficients.

Jennifer responded,

Do you mean if c is 0 and d is 1 that would mean (ax2 + b)(0x2 + 1) = ax2 + b?

She sees the obvious polynomial that would work as an identity if we were using ordinary polynomial multiplication, namely the polynomial 1. So she clearly has the idea of an identity. But there are two problems: These aren’t just any polynomials, and the “multiplication” is not ordinary multiplication! Doctor Fenton replied:

First of all, the identity e(x) must be an element of P2ˆ(x).  Is 0x2 + 1 an element of this set?  What were the requirements that e(x) = cx2 + d be in P2ˆ(x)?

Perhaps it is confusing to call this way of combining two elements of the set a product, since although the elements of the set are written as polynomials, they don’t have the usual properties of polynomials.  The given rule

(ax2 + b)*(cx2 + d) = (ac)x2 + (ad + bc)

is just a way to take two elements of the set, and create a third element of the set.  I’ll call this a “*-product” (read “star-product”) of the two elements.

What is the product (ax2 + b)*(0x2 + 1), according to the given rule for the *-product?

(One thing that seems to be taken for granted in this problem, is that two elements p(x) = ax2 + b and q(x) = cx2 + d are equal if and only if a = c and b = d.  You need to use that fact.)

In some groups, equality is defined differently; I’ll be giving an example at the end. Here, polynomials are considered equal if they are the same polynomial, as usual.

The key step that hasn’t been actually done yet is to write out what an inverse is in terms of the definition of the operation. I imagine he is giving Jennifer a chance to figure this out on her own, which will help her internalize it.

Jennifer answered

I am so sorry but I just don’t understand. Maybe the first thing I don’t understand is what a star product is. I understand what an identity element is. I understand it is what maintains the original element. In addition it would be zero, in multiplication it would be one.

She is indeed understanding what an identity is. If the operation were addition, the identity (called the additive identity) would be 0, since \(a+0=a\) for any number a; and if the operation were multiplication, identity (the multiplicative identity) would be 1, since \(a\cdot 1=a\) for any number a. The new thing, that is taking time to absorb, is the very idea of a made-up operation.

An hour later she added several good ideas:

So… If (ac)x2 has to equal ax2 would c not be one? And if ad needs to = a, would that mean d needs to equal 1 as well?  Or am I still confused? I just realized that formula was the definition of a star product.

Although this is not quite right, she has seen the central issue: She has to use the definition of the operation in order to see what the identity element will be.

Doctor Fenton replied:

You are partly right.  Since the product p(x)*e(x) must equal p(x), then

p(x)*e(x) = (ax2+b)*(cx2+d)   must equal    p(x)=ax2+b.

I still haven’t seen you write out the *-product p(x)*e(x).  (If you don’t like that name, just call it the product, but it is not the product in the usual sense of polynomials.  It is a binary operation on these polynomial-looking elements of P2ˆ(x): you give me p(x) and q(x), and I use them to make a third element using the recipe

(ax2 + b)*(cx2 + d) = (ac)x2 + (ad + bc)

This *-product must be the same as ax2+b, so by the equality rule I mentioned earlier (which the problem seems to just assume), then you are correct that ac must equal a, and so c must equal 1.  But then you say

And if ad needs to = a, would that mean d needs to equal 1 as well?

Where did you see that ad must equal a?  Looking at the formula for the *-product, the constant term is ad + bc, and so ad + bc must equal b.  If c = 1, what value of d will make ad + bc=b?

If the identity is \(cx^2+d\), then the definition of identity means that, for any \(ax^2+b\), $$(ax^2+b)*(cx^2+d)=ax^2+b$$ that is, $$(ac)x^2+(ad+bc)=ax^2+b$$ What must c and d be?

Jennifer did just this, but using u and v instead of c and d (likely having seen something similar in an example she had):

So… (ax2 + b)(e) = ax2 + b

By definition with * product (ax2 + b)(ux2 + v) = (au)x2 + (av + bu)

With the identity I want (ax2 + b)(ux2 + v) = ax2 + b

Then av + bu=b

If u = 1 then av + b(1) = b, which would only work if v = 0

Which would mean there is an identity element where e = (ux2 + v)

She has omitted the “*” in her multiplications, which makes what she wrote a little confusing, but she did the right things. Here is her work written a little more correctly: $$(ax^2+b)*(ux^2+v)=ax^2+b\\ (au)x^2+(av+bu)=ax^2+b\\ au=a, av+bu=b\\au=a\Rightarrow u=1\\ av+b(1)=b\Rightarrow av=0\Rightarrow v=0$$

Doctor Fenton just had to state the conclusion:

Yes, but put in the values of u and v you found, so that the identity element is e(x) = 1x2 + 0 = x2 .  Now you are ready to tackle the inverse. If p(x) = ax2 + b, can you find an element q(x) = cx2 + d such that p(x)*q(x) = e(x)?

So the identity element is \(x^2\); this works as an identity because, for any \(p(x) = ax^2+b\), $$p(x)*e(x) = (ax^2+b)*(1\cdot x^2+0)= (a\cdot 1)x^2+(a\cdot 0+b\cdot 1)=ax^2+b=p(x)$$

Finding the inverse of an element

Jennifer took on the new challenge (again omitting the *, but otherwise correct):

For inverse,

(ax2 + b)(cx2 + d) = x2

so (ac)x2 + (ad + bc) = x2

so c = a-1 and

ad + bc = 0

ad + ba-1 = 0

ad = -ba-1

Then I am stuck.  I know that I can multiply both right sides by a to eliminate the a-1 on the right side but I am not sure how that helps me

She is very close! If \((ax^2 + b)*(cx^2 + d) = x^2\), then \((ac)x^2 + (ad + bc) = 1x^2+0\), so that \(ac=1\) and \(ad+bc=0\), so \(c=\frac{1}{a}\) and \(ad+b\frac{1}{a}=0\). She just needs a little nudge …

Doctor Fenton:

Remember what you are trying to do: p(x) is given, so you know a and b, and you know that a ≠ 0.  You are trying to FIND c and d.  You found c.  How can you find d from the equation ad = ba-1?

Jennifer repeated the same work, going a little farther:


ad + bc = 0

ad = -bc

ad = -ba-1

a-1 a d = -a-1ba-1

d = -a-1 b a-1

These are not commutative, correct? So can I simplify beyond this?

Now she has absorbed a little too well the idea of abstract algebra! She is forgetting that at this point she is doing ordinary algebra on ordinary numbers. This, too, is not uncommon when first learning these ideas.

Doctor Fenton reminded her where she was:

But a, b, c, and d  are real numbers, and the arithmetic used (ac and ad + bc) is ordinary multiplication and addition of real numbers, which ARE commutative.  In fact, the group is commutative: p(x)*q(x) = q(x)*p(x).

So, now you can write the inverse q(x) of p(x) = ax2 + b.

(We hadn’t mentioned until now that the star operation is commutative; we don’t need to know this, but it was assumed when we defined the identity and the inverse without doing the operation on both sides. To see this, observe that $$p(x)*q(x) = (ax^2+b)*(cx^2+d) = (ac)x^2+(ad+bc)$$ while $$q(x)*p(x) = (cx^2+d)*(ax^2+b) = (ca)x^2+(da+cb) = (ac)x^2+(ad+bc)$$ These are equal, so the operation is commutative. There is a certain symmetry in the operation’s definition that makes this work, which is so obvious to those accustomed to it that it almost goes without saying – and almost did!)

Jennifer finished the work:

Then d = -b/a2

or d = -ba-2

So the inverse of \(ax^2+b\) is \(a^{-1}x^2-ba^{-2} = \frac{1}{a}x^2-\frac{b}{a^2}\).

Doctor Fenton:

That’s right! Good work!

Is this making more sense now?  Math is sometimes thought of as an intellectual game, in which one can make up the rules, and this problem is an example.  However, much, if not most, of the time, the rules are given by a situation, and seeing a structure such as a group can tell you things you wouldn’t normally realize.  The idea of a group is very important in physics, for example.

Proving closure

Jennifer confirmed the central reason for her confusion, and continued to the fact we skipped over, namely closure:

Yes!  Thank you so much! I am in abstract algebra but haven’t really taken math since Calculus 1, 25 years ago. I have found that I am missing a lot of information. I thought star product was regular multiplication and I was so confused.

For the closure part of the proof can I just say that since it is given that a, b are real numbers and p, q are real numbers then p * q would also include real numbers and therefore be closed under multiplication or do I need more information?

No, it takes a little more than that. Doctor Fenton gave a reminder:

Remember that the two conditions for p(x) = ax2 + b to be in P2ˆ(x)  were that a and b be real numbers, and a ≠ 0.  You should show those properties for the product of two such elements.  The product has ac and ad + bc.  Those are real numbers.  Is ac non-zero?

The reason that the two coefficients in the “product” are real numbers is that addition and multiplication of real numbers are closed (or rather, that the real numbers are closed under ordinary addition and multiplication). We just need to see that the coefficient of \(x^2\) in the “product” is non-zero. Is it always so?

Jennifer took a stab at it:

So… because we determined that ac = 1. Therefore a does not equal 0 and c does not = 0  so it is closed

Doctor Fenton corrected this reference to the wrong previous work,

No, ac doesn’t have to be 1, it can be any non-zero real number.  For example, πx2 + (ln 2) and (√2)x2 – 1 are both elements, and their product is (π(√2) x2 +(-π + (√2)(ln 2)), so ac need not be 1.

Jennifer made the required small fix:

Right. That was for the identity. Just when I think I have it. Then if a does not equal 0 and c does not equal 0 then ac does not equal 0 and it is closed.

Doctor Fenton confirmed the answer:


Jennifer closed with thanks:

Thank you so much for sacrificing your Easter to help me!! May God richly bless you.

A couple extras

Now let’s get back to a couple things I mentioned at the start.

Proving the operation is associative

The professor had already shown the that operation is associative, so we didn’t get to see that. Let’s fill in the gap, for our own sake.

The operation “*” is associative if, for any three elements p, q, and r, it is true that \((p*q)*r = p*(q*r)\). So we just have to compute those.

$$(p(x)*q(x))*r(x) = \left((ax^2+b)*(cx^2+d)\right)*(ex^2+f) =\\ \left((ac)x^2+(ad+bc)\right)*(ex^2+f) =\\ ((ac)e)x^2+((ac)f+(ad+bc)e) =\\ (ace)x^2+(acf+ade+bce)$$

$$p(x)*(q(x)*r(x)) = (ax^2+b)*\left((cx^2+d)*(ex^2+f)\right) =\\ (ax^2+b)*\left((ce)x^2+(cf+de)\right) =\\ \left((a(ce))x^2+(a(cf+de)+b(ce))\right) =\\ (ace)x^2+(acf+ade+bce)$$

These are equal, so we have it.

This is (almost) a familiar group in disguise!

I mentioned something I saw from the start that made some of the answers obvious to me. What was it?

Compare the definition of  “*”, $$p(x)*q(x) = (ax^2+b)*(cx^2+d) = (ac)x^2+(ad+bc)$$

to this definition of addition of fractions: $$\frac{b}{a}+\frac{d}{c} = \frac{ad+bc}{ac}$$

It appears that the operation works like adding fractions, if we equate \(ax^2+b\) to \(\frac{b}{a}\). That is, we can think of \(a\) as the denominator of a fraction and \(b\) as the numerator, and the operation as addition. And the fact that a can’t be zero agrees with the fact that a denominator can’t be zero/

Then since the additive identity for fractions is \(0 = \frac{0}{1}\), we can expect the identity for our operation to be \(1x^2+0\).

And since the additive inverse of \(\frac{b}{a}\) is \(-\frac{b}{a}\), the inverse of \(ax^2+b\) should be \(ax^2-b\). Well, not quite! As we saw, it’s \( \frac{1}{a}x^2-\frac{b}{a^2}\), which corresponds to the fraction \(\displaystyle\frac{-\frac{b}{a^2}}{\frac{1}{a}}\); this is equivalent to \(-\frac{b}{a}\) when you multiply numerator and denominator by \(a^2\).

What happened? Recall the comment that they implicitly take two elements as equal when corresponding coefficients are equal? In fractions, \(\frac{a}{b}=\frac{c}{d}\) not only when \(a=c\) and \(b=d\), but whenever \(ad=bc\). So in fractions, we are treating some different ordered pairs \((a,b)\) as equal, while in the group under discussion, they are treated as distinct. So our group is not isomorphic to the group of fractions under addition (that is, it doesn’t have all the same behavior), but it is closely related.

Fractions can be technically defined as ordered pairs \((n,d)\) of real numbers such that d is non-zero, fractions are considered equal as stated above, and operations are defined as I showed for addition. The group in this problem can similarly be thought of as ordered pairs \((a,b)\) with a few conditions and an operation defined. The fact that they were written as polynomials was not essential to the problem, but in fact disguised their similarity to fractions.

2 thoughts on “Proving Certain Polynomials Form a Group”

  1. Pingback: Proving Two Groups are Isomorphic – The Math Doctors

Leave a Comment

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.