Group Theory
Nilpotent Number
Characterizations of the set of positive integers $n$ such that every group of order $n$ is (i)
cyclic, (ii) abelian, or (iii) nilpotent.
Say that a positive integer $n > 1$ is a **nilpotent number** if $n = p_1^{a_1} \cdots p_r^{a_r}$
(here the $p_i$'s are distinct prime numbers) and for all $1 \leq i,j \leq r$ and $1 \leq k \leq
a_i$, $p_i^k \not \equiv 1 \pmod{p_j}$. Also, let us say that $1$ is a nilpotent number.
(So, for instance, any prime power is a nilpotent number. A product of two distinct primes $pq$ is a
nilpotent number unless $p \equiv 1 \pmod q$ or $q \equiv 1 \pmod p$.)
Then, for $n \in \mathbb{Z}^+$:
- (Pazderski, 1959) Every group of order $n$ is nilpotent iff $n$ is a nilpotent number.
- (Dickson, 1905) Every group of order $n$ is abelian iff $n$ is a cubefree nilpotent number.
- (Szele, 1947) Every group of order $n$ is cyclic iff $n$ is a squarefree nilpotent number. and n is cyclic iff $gcd(n,\phi(n))=1$
- Integral Domain in $R$
- Every Ideal is a Principal Ideal Ring i.e { Every Ideal can be written as I = $\left \langle a \right \rangle $ = { $ra$ | $r \in R $}
Principal Ideal Domain (PID)
- Integral Domain in $R$
- Every Ideal is a Principal Ideal Ring i.e { Every Ideal can be written as I = $\left \langle a \right \rangle $ = { $ra$ | $r \in R $}
Factor Ring
If all this seems vile and abstract, fear not: quotient rings are simple and quite fun to work with
in
practice. We’ve already seen one example: the integers mod n. Take R = Z and I = (n). Then
R/I = Z/nZ, and this is just the usual integers mod n where we do all thee addition and
multiplication
mod n.
Now let’s look at quotients of polynomial rings, as this is the fun bit. We’ll just do examples
until
you get the idea:.
Examples
- Take R = C[x], and I = (x − 2). We said before that the coset I acts as the zero element in R/I, so x − 2 = 0. This is the same, using the addition rule for cosets, as saying that x = 2. So how does R/I look? Well, it’s just like the polynomial ring, except we’ve replaced x with 2 and put bars over everything. Since 2 is already in C, we haven’t really added anything, so in fact this quotient ring is just like C..
- Take R = R[x], I = (x 2 + 1). In the quotient ring R[x]/(x2 + 1), we have cosets of polynomials, but with the rule that x 2 + 1 = 0, which we rewrite as x 2 = −1. So we think of elements of R/I here as polynomials, but with the extra rule that x2 = −1. This is something we’ve seen before: it’s basically the same as C, where we take R and adjoin a square root of −1, which in the current setting is being denoted by x..
- Let’s work backwards a bit. Say you want to work with a number system that’s like the rational numbers, but also for some reason includes the element √ 2. We can build this number system, which before we’ve called Q[ √ 2], using quotient rings. Start with Q. We’ll need to adjoin a variable (call it x), and then set this variable equal to √ 2. We cannot just quotient out by(x − √ 2). because that’s not a polynomial with rational coefficients. So the correct quotientring seems like it should be Q[x]/(x2 − 2). So the ring Q[x]/(x2 − 2) is like the rationals, butwith an extra symbol x, which we know should behave like √2, namely x2 = 2. We can say thatQ[x]/(x2 − 2) ∼= Q[√2], where the latter is to be thought of as a subring of R (or C).4. Let R = R[x] and I be the ideal generated by x3 − 1. Now I’m going to stop writing the barson top of everything, for simplicity. So we think of elements of this ring as polynomials in x, butsubject to the relation x3 = 1. What is (x4 − x2) times (x 2 + x + 1) in this ring? It doesn’t have degree 6, like you’d expect in a normal polynomial ring. First we can simplify the first one to x − x 2 using the relation, and then multiply them out to give (x − x 2 )(x 2 + x + 1) = x 3 + x 2 + x − x 4 − x 3 − x 2 = x − x 4 , and since x 3 = 1, x 4 = x, so this is just zero! So the first thing we notice is that this ring has zero divisors - we just found two of them - x 4 − x 2 and x 2 + x + 1. There is a more enlightening way to see that their product is zero, however, by diligently factoring everything first: (x 4 − x 2 )(x 2 + x + 1) = x 2 (x + 1)(x − 1)(x 2 + x + 1) = x 2 (x + 1)(x 3 − 1). 16 This shows that their product is a multiple of x 3 −1, hence lies in I. So (x 4 −x 2 )(x 2 +x + 1) and x 3 − 1 are equivalent, hence the coset [(x 4 − x 2 )(x 2 + x + 1)] is the same as the coset [x 3 − 1], which is just I, and we know that this functions as the zero element in the quotient ring. Notice that the zerodivisor phenomenon basically happened because the generator for I factored: x 3 −1 = (x −1)(x 2 +x + 1). That means right away that the two polynomials x −1 and x 2 +x + 1 are zerodivisors. We’ll come back to this when we discuss prime ideals. 5. Here’s a cool example that shows how you can do calculus without limits. Start with the ring R[x], and adjoin a new element , giving the ring R[x][] = R[x, ]. Finally, take the quotient by the ideal ( 2 ), this forces to satisfy the relation 2 = 0. Let R = R[x, ]/( 2 ) be this new ring. We can compute derivatives in this ring as follows: pick a polynomial whose derivative you want to compute, say f (x) = x 3 . Now look at f (x + ) = (x + ) 3 = x 3 + 3x 2 + 3x 2 + 3 = x 3 + 3x 2 . All the higher terms disappear because of the relation 2 = 0, and the coefficient of the remaining term is just 3x 2 , the derivative! In a HW exercise you will be asked to prove that in general, f (x + ) = f (x) + f 0 (x) If you work instead with the relation n = 0, when you expand f (x +), you will get the first n + 1 terms of the Taylor series for f .