What you'll learn

Groups and polynomials. We are all familiar with the quadratic formula, which gives an explicit

description of the roots of a polynomial of degree 2, say with real coecients:

ax2 + bx + c = 0 =) x =

􀀀b 

p

b2 􀀀 4ac

2a

The key observation for us is that this expression for the roots involves only the coecients of

the given polynomial and some basic algebraic operations: addition, subtraction, multiplication,

division, and taking a square root. If we go back a degree, it is also true that the roots of a linear

polynomial can be expressed in terms of its coecients and basic algebraic operations (only division

is needed in this case!):

ax + b = 0 =) x = 􀀀

b

a

:

The search for analogous formulas for polynomials of higher degree is a problem which dates

back centuries, even millennia! The cubic formula for polynomials of degree 3 is much more

complicated than the quadratic formula above (it requires higher-order root extractions), and the

quartic formula for polynomials of degree 4 even more so (you can see what they look like on

Wikipedia!), but the point is that such formulas exist. However, in the 19th century it was proven

that no analogous quintic formula for polynomials of degree 5 existed. This seems quite surprising

at rst, since it is not all all clear what breaks down when we make the jump from degree 4 to

degree 5. It turns out, as we'll see fully in the spring, that the reason for this has to do with the

structure of the \group of permutations" of the roots of such polynomials.

We will avoid giving any formal denitions for now, but here is the basic (and at this point quite

vague) idea. Consider the polynomial x2 􀀀 2 with roots

p

2;􀀀

p

2. There are two possible permu-

tations of the roots in this case|do nothing, or exchange them|which thus form a \permutation

group" with two elements. The polynomial x3 􀀀2 has three roots: 3

p

2 and two complex conjugate

roots. There are 3! = 6 ways of permuting these 3 roots, and so we get a group of permutations

with 6 elements in this case. Now consider (x2􀀀2)(x2􀀀3). This has roots

p

2;􀀀

p

2;

p

3;􀀀

p

3, and

there are 4! = 24 ways of permuting these. However, in this case it turns out that not all of these

24 possible permutations should actually be allowed; the issue is that permuting, say,

p

2 and

p

3

exchanges roots of the two factors x2 􀀀 2 and x2 􀀀 3, and, for reasons we will leave for the spring,

this should not actually be a valid permutation. We only get four allowable permutations|the two

coming from permuting 

p

2 and the two which permute 

p

3|and thus a group with 4 elements.

What's Included

• 27 Lectures
• 1 Assignments
• 1 downloadable resources
• Access on tablet and phone
• Certificate of completion

Requirements

• --> The upshot is that the question as to whether or not we can express the roots of a polynomial in a certain way is intimately related to properties of these groups of root permutations, and that properties of these groups reflect properties (or “symmetries”) of the roots. To give an answer as to why a quintic formula does not exist—although not an answer which will make any sense as this point—the fact is that there exist polynomials of degree 5 which have the so-called alternating group A5 as its group of root permutations, and A5 has the property of being a simple, non-abelian group; it is this property which prevents there from being a nice way of expressing the roots of such a polynomial. (All groups which arise from polynomials of degree 4 or less—one example being the alternating group A4—are either non-simple or simple and abelian, so there is nothing which stands in the way of there being a nice way of expressing the roots.) We will soon understand what all of these terms above mean, but we mention this now in order to give some motivation for the study of groups.