Question

prove that R* x R is a group under the operation * defined by (a,b) * (c,d) = (ac, bc + d)

Answer 1

is it \((a, b)\circ (c,d)=(ac, bc+d)\) ?

Answer 2

and i guess we need to know what \(R^*\) means in this context
i take it \(R\) is a ring?

Answer 3

R is all real numbers

Answer 4

R* is a multiplicative group of all
real numbers

Answer 5

R* is the nonzero reals

Answer 6

i think the biggest challenge will be to find the inverse
first figure out what the identity is

Answer 7

(a,b) * (c,d) = (ac, bc + d)

(ac, bc + d) * (x,y) = (acx, bcx+dx+y)

is this associative?

Answer 8

i hate checking associativity, it is a pain
used to get away with saying "in inherits it from the original group" but not in this case

Answer 9

(c,d) * (x,y) = (cx, dx + y)

(a,b) * (cx, dx + y) = (acx, bcx + dx+y)

yay! its associable

Answer 10

whew

Answer 11

(a,b) * (x,y) = (a,b) is identity

Answer 12

now for the identity, solve
\[(x,y)(c,d)=(c,d)\] for \((x,y)\) you get
\[(x,y)(c,d)=(xc,yc+d)=(c,d)\implies xc=c\implies x=1\] and \[yc+d=d\implies y=0\]

Answer 13

identity is \((1,0)\) now for inverses

Answer 14

can we assume commutability in groups?

Answer 15

no

Answer 16

but we could check to see if it is commutative i guess
hmmm

Answer 17

oh seems unlikely since it is not symmetric in the second coordinate

Answer 18

and identity elements needs to be unique, correct?

Answer 19

biggest job i think it to find \((a,b)^{-1}\)

Answer 20

yeah identity is always unique

Answer 21

doesnt it only have to be commutative for abelian groups?

Answer 22

commutative is the definition of abelian yes

Answer 23

i doubt this is commutative
you still have to show it contains inverses

Answer 24

I mean from what you have already done, haven't you proved that is it a group?

Answer 25

no
amistre showed it was associative
i found the identity element
you still have to show it contains inverses

Answer 26

and i guess you have to show it is closed under the operation too

Answer 27

(a,b)*(1,0) = (a1,b1+0) = (a,b)

(1,0)*(a,b) = (1a, 0a + b) = (a,b)

Answer 28

since R*xR does not contain the (0,n) element, we would have to see of this is possible

Answer 29

well, since 0 is not a first element in any of the cases, its closes

Answer 30

guess this is a good time to review the axioms
1) set must be closed under the operation
2) the operation must be associative
3) there must exist and identity element \(e\) such that \(ex=xe=x\) for all \(x\in G\)
4) it must contain inverses, i.e. if \(x\in G\) there must exist an \(x^{-1}\) with
\[x^{-1}x=xx^{-1}=e\]

Answer 31

we got 1 through 3, now need 4

Answer 32

(a,b)*(x,y) = (1,0) :)

Answer 33

since we got reals and no zero firsts, ax = 1 is possible

Answer 34

bx+y = 0 when x=-y/b is possible

Answer 35

hmm, for b not= 0 tho

Answer 36

theres no reason why y=-bx is not possible tho

Answer 37

in fact it is \(y=-\frac{b}{a}\) i think

Answer 38

\[(a,b)*(\frac1a,-\frac ba)=(\frac aa,\frac ba-\frac ba)=(1,0)\]

Answer 39

sin ce a in R*, a not= 0

Answer 40

is that commutative? :)

Answer 41

\[(a,b)(x,y)=(ax,bx+y)=(1,0)\]
\[ax=1\implies x=\frac{1}{a}\]
\[bx+y=0\implies \frac{b}{a}+y=0\implies y=-\frac{b}{a}\] so inverse of \((a,b)\) is
\[(a,b)^{-1}=(\frac{1}{a},-\frac{b}{a})\]

Answer 42

oh i am a little behind i see

Answer 43

must be the weather up there

Answer 44

the inverse as also communicable

Answer 45

still g od da mn cold!!

Answer 46

been nice here in florida :)

Answer 47

we had snow yesterday...

Answer 48

yeah will i will take to you in july when it is 102

Answer 49

lol, 102 is just a calm pleasant day :)

Answer 50

i appreciate all the help today! get excited for next week when I study for finals. ;)

Answer 51

well this was a fun exercise, time to do some work
later all

Answer 52

my teacher gave me a real analysis and an abstract algebra book to read over the summer

Answer 53

I'm taking analysis 1 and 2 next year. get excited for that too

Answer 54

@urbanderivative i hope it is clear what you have to do to solve this
write down the axioms
check that it satisfies each one

Answer 55

@amistre64 what algebra book?

Answer 56

.... something old, prolly from when she was in college 20 years ago. i cant recall the title at the moment

Answer 57

so twenty years ago is old??

Answer 58

:) at todays ages, yes

Answer 59

if you can find a copy of "herstein" topics in algebra, i think that is still the gold standard, although i guess it is really really old

Answer 60

maybe from the 60's although how we call that "mid century"

Answer 61

or 'mad men era'