Question

let G be the set of all polynomials of the form ax^2+bx+c with coefficients from the set {0, 1, 2}. we can make G a group under addition by adding the polynomials in the usual way, except that we use modulo 3 to combine the coefficients. with this operation, prove that G is a group of order 27 that is not cyclic

Answer 1

Answer: All the statements can be proved using definitions from group theory.

Step-by-step explanation:

We will use the fact that the set $\{0,1,2\}:=\mathbb{Z}_3$ is a group under addition modulo 3 (denoted by +). This is because addition modulo 3 is associative, 0 is the identity element for this group and -0=0, -1=2, -2=1.

Let $ax^2+bx+c,dx^2+ex+f, gx^2+hx+i \in G$.

Addition in G is well defined, because $ax^2+bx+c+dx^2+ex+f=(a+d)x^2+(b+e)x+(c+f)$ and the coefficients of this last polynomial are computed using addition modulo 3, which is well defined.

Because addition modulo 3 is associative, we have that $(ax^2+bx+c+dx^2+ex+f)+gx^2+hx+i=((a+d)x^2+(b+e)x+(c+f))+gx^2+hx+i=((a+d)+g)x^2+((b+e)+h)x+((c+f)+i)=(a+(d+g))x^2+(b+(e+h))x+(c+(f+i))=ax^2+bx+c+dx^2+((d+g)x^2+(e+h)x+c+(f+i))=ax^2+bx+c+dx^2+(dx^2+ex+f+gx^2+hx+i)$ therefore the addition defined in G is associative.

The identity element for this group is the polynomial $e=0:=0x^2+0x+0\in G$. Indeed, we have that  $ax^2+bx+c+0=(a+0)x^2+(b+0)x+(c+0)=ax^2+bx+c=(0+a)x^2+(0+b)x+(0+c)=0+ax^2+bx+c$

The inverse element of a polynomial $ax^2+bx+c \in G$ is the polynomial $(-a)x^2+(-b)x+(-c) \in G$, because $ax^2+bx+c+(-a)x^2+(-b)x+(-c)=(a-a)x^2+(b-b)x+(c-c)=0=(-a)x^2+(-b)x+(-c)+]ax^2+bx+c$

This group has order 27: to count all the polynomials in G we have to count the number of ways to choose a, b and c and construct a polynomial. a,b and c can be either 0,1 or 2 so each one can be chosen in 3 ways. In total, there are 3×3×3=27 possible choices that result in a element of G.

This group is not cyclic. Take any polynomial $ax^2+bx+c \in G$, then its generated group is the set $<ax^2+bx+c>=\{0,ax^2+bx+c,2ax^2+2bx+2c\}$. This is because any element of the form$n(ax^2+bx+c)=nax^2+nbx+nc, n\in \mathbb{Z}$ can be reduced to $(n\mod 3)ax^2+(n\mod 3)bx+(n\mod 3)c$ and $n\mod 3$ only takes the values 0,1,2. The generated group of any element of G has less than 27 elements, so G can't be cyclic.