Groups Questions

[Loktart]

I tried to play around with the identity and the axioms given at the end, however, I keep going round in circles proving that e = e =(

Here is the problem

Let G = (G,., e) be a group. Prove carefully that for all a in G, for all b in G,
there exists a unique x in G such that
ax = b:
Does this x also necessarily satisfy xa = b? If yes, prove that it does. If not, give
a example of a G for which it will be not be true that ax = b => xa = b.

Things you may assume.
A natural number p is prime if and only if 8a 2 Z; 8b 2 Z; pjab ) pja or pjb.
The axioms for a group: A group consists of (G, . , e) in which G is a set,
. : G X G -> G is a function, and e in G an element such that:
(1) For ALL a in G, For All b in G, For All c in G, (a . b) . c = a . (b . c). (Associativity)
(2) For ALL a in G, e . a = a . e = a. (Identity)
(3) For ALL a in G, There Exists b in G; ab = ba = e (Existence of inverses).

 

[daon]

First, existence:

\(\displaystyle \text{Let } a,b \in G\).

\(\displaystyle b=eb = (aa^{-1})b = a(a^{-1}b)\)

\(\displaystyle \text{So, } x=(a^{-1}b) \in G \text{ satisfys our requirment.}\)

Now, uniqueness:

\(\displaystyle \text{Let }b=ax=ax'.\)

\(\displaystyle \text{Then }a^{-1}(ax)=a^{-1}(ax')\)

Can you finish?

For the second part, do we necessarily know that

\(\displaystyle ab=ba \,\, \forall \,\, a,b \in G\)?

What property is this called? Do you know of any groups not having this property?

 

[Loktart]

(a^-1)(ax)=(a^-1)(ax')

=> ((a^-1)(a))x= ((a^-1)(a))x'
=> x=x'

I would assume for the 2nd part is true since its associativity essentially.

 

[daon]

Associativity allows us to say the following:

\(\displaystyle a(bc) = (ab)c\).

That is not the same thing as what I wrote.

 

[altt-gaming]

Does it mean that x=x', or rather, ex = ex'

 

[daon]

Yeah, that was correct.

By the "second part" I meant where it asked if b=ax implies b=xa.

 

[Loktart]

if x= (a^-1)(b)

ax=b = (a^-1)(b)(a)=> b=b

then xa=b = (a^-1)(b)(a)=b => b=b

is it that obvious or am I missing something?

 

[daon]

then xa=b = (a^-1)(b)(a)=b => b=b

is it that obvious or am I missing something?

That is not true, which is my point.

 

[Loktart]

Let b=ax=ax'

ax'=b
a(a^-1)(b)=b

then x'=(a^-1)(b)=x

thus uniquness.

Not all groups are cumatitive, thus,

ax=b = xa=b only if your group is cumatitive.

a possible group that isn't cumatitive, is the D4 symmetry of square group, where Fh X Fd not equal to Fd X Fh

Thank you for the help.