# Groups in my algebra class

1. Oct 8, 2003

### wubie

Hello,

I just started doing groups in my algebra class and I am struggling with the abstraction of it as usual.

Here is how my class defines a group:

A group is a nonempty set G with a binary operation "o" such that for all of x,y,z which are elements of the group the following holds:

(P0) If x and y are elements of G, then x o y are elements of G.
(P1) x o (y o z) = (x o y) o z
(P2) There's a u which is an element of G such that u o x = x = x o u
(P3) For every x which is an element of G there is an x-1 such that x o x-1 = u and x-1 o x = u.

Here is my question:

iii) The set of all rational numbers except -1, with the operation o defined by x o y = x + y + xy.

iv) The set of all integers that are multiples of a fixed integer d, with the operation of addition.

I think I have iv). I believe it is a group.

By (P0), let t = xd and u =yd where x and y are elements of Z. Then u + t = xd + yd = d(x + y).

By (P1) let p = wd, t = xd and u =yd. Then p + (t + u) = (p + t) + u

By (P2) let u = 0*d and t = xd then t + u = 0*d + xd = 0 + xd = xd and vice versa.

By (P3) Let t = xd and w = -xd then t + w = u = w + t.

Now for iii), I am not sure if I understand the situation.

I think that addition by rationals results in rationals. I also think that addition of rationals are associative. But I am not sure how to apply P2 and P3 especially since I am not sure how to interpret

and how to apply the axioms P2 and P3 to it.

It may seem obvious to others, but it is not obvious to me (sometimes I have to read the most elementry passages many times and wait for a period of time before I absorb it). Perhaps someone who understands the situation can reword it for me? It may help me understand it better.

Any help is appreciated. Thankyou.

Last edited by a moderator: Feb 6, 2013
2. Oct 8, 2003

### Hurkyl

Staff Emeritus
x o y is just infix notation for an arbitrary binary function. The group axioms could be rewritten as:

A group is a nonempty set G with a binary function f such that for all x, y, z in G:

f(x, y) is in G
f(x, f(y, z)) = f(f(x, y), z)
there exists a u such that f(u, x) = x = f(x, u)
For every x there is a w such that f(x, w) = u = f(w, x)

And for problem iii), f(x, y) is defined to be x + y + xy

The infix notation is typcially used because it is easier to read and emphasizes the analogy between group/ring operations and ordinary addition/multiplication.

3. Oct 8, 2003

### HallsofIvy

Staff Emeritus
To show that a set with a given operation is a group you must show that each of the "rules" is satisfied.

Yes, it is true that the sum of two rational number is a rational number and that the product of two rational numbers is a rational number (just look at a/b+ c/d and (a/b)(c/d)) so the rationals with "x+ y+ xy" is closed under this operation.

To determine if there exist an identity (your "there exists a u such that f(u, x) = x = f(x, u)") you need to look at x+ u+ xu= x. Can you solve for u? Is it independent of x?

To determine if each member has an inverse (your "For every x there is a w such that f(x, w) = u = f(w, x)") you need to look at
x+ w+ xw= u (u is the number you determined above. Of course, if there was no such number you are finished.). Can you solve for w? (It will, in general, depend on x.)

The "associative law" (x o (y o z) = (x o y) o z) is typically the hardest thing to show.

xo (yoz)= xo(y+ z+ yz)= x+ (y+z+ yz)+ x(y+ z+ yz)

(x o y)o z= (x+ y+ xy)o z= (x+ y+ xy)+ z+ (x+ y+ xy)z

Are those the same?

4. Oct 9, 2003

### wubie

Thank you Hurkyl and Hallsofivy.

That was exactly what I needed to finish the question.

Cheers.