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 Exercise [13.01] 
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Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Exercise [13.01]
We are given a set G which is closed under the given binary operation(juxtaposition) such that the following axioms hold:
1.a(bc)=(ab)c for all a,b,c in G.
2.An element e exists in G such that ea=a for all a in G.Left identity
3.There exists a left inverse a' in G such that a'a=e for every element a in G.Left inverse
We need to prove that G satisfies the original group axioms.
Now,
ex=x for all x in G.
Therefore, ee=e. ......(1)
Now, a'a=e ......(2)
From (1) and (2),
(a'a)e=e ........(3)
From axiom 3 it follows that every element in G has a left inverse,which implies
a''a'=e for an a'' in G.
Pre-Multiplying b/s of (3) by a'' we have
a''(a'a)e=a''e
or,(a''a')ae=a''e(associative law)
or, ae=a''e (as a''a'=e) ........(4)
Also,a''e=a''(a'a)=(a''a')a=ea ........(5)
From (4) and (5)
ae=ea=a(from axiom 2) ........(6)
Now,ae=a''e
or,ea=ea'' (from (6))
or,a=a''
Thus,
e=a''a'=aa'
which together with a'a=e implies
a'a=aa'=e
Hence,G is a group under the given operation.
On the other hand the existence of a left identity and a right inverse seems to be insufficient because the above proof breaks down midway between equation 3 and 4.


22 Aug 2008, 14:47

Joined: 13 Aug 2009, 00:08
Posts: 13
Post Re: Exercise [13.01]
I would do it this way: Assume from Penrose that:
\forall a : 1a = a; a^{-1}a = 1

Then we also have b^{-1}b=1 for any b, so let's use b=a^{-1} (from Penrose's hint).

Then consider aa^{-1}:
aa^{-1} = ab = 1ab = b^{-1}bab = b^{-1}a^{-1}ab = b^{-1}1b = b^{-1}b = 1
(with free use of the distributive law).

We then also have
a1 = a(a^{-1}a) = (aa^{-1})a = 1a = a

Done.

As for the second part, I would have thought the only way really to do that would be to find a counter-example. Anyone?


25 Aug 2009, 21:45
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [13.01]
flashbang wrote:
Then consider aa^{-1}:
aa^{-1} = ab = 1ab = b^{-1}bab = b^{-1}a^{-1}ab = b^{-1}1b = b^{-1}b = 1
(with free use of the distributive law).

As for the second part, I would have thought the only way really to do that would be to find a counter-example. Anyone?

For the second part, given that now a1=a, you would have to write:

aa^{-1} = ab = (a1)b or
aa^{-1} = ab = a(b1)

Neither of which leads to aa^{-1}=1 as far as I can see!


26 Aug 2009, 09:42
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [13.01]
Another way is to say:
Given that 1a=a, and a^{-1}a=1, show that a1=a and that aa^{-1}=1.
So to prove that a1=a, let's ask what b satisfies ab=a?
So ab=a
\Rightarrow a^{-1}(ab)=a^{-1}a
\Rightarrow (a^{-1}a)b=1
\Rightarrow 1b=1
\Rightarrow b=1
\Rightarrow a1=a

To prove that aa^{-1}=1 let's say that since a^{-1} is, by definition, a member of the group it too must have an inverse, say c.
Then from the definition of the inverse
ca^{-1}=1

So (ca^{-1})a=1a
\Rightarrow c(a^{-1}a)=a
\Rightarrow c1=a
\Rightarrow c=a since c1=c by the proof above.
So aa^{-1}=1

However, if we start with a1=a and a^{-1}a=1 then we can't prove that 1a=a and so, since the second line of the second proof above depends on this fact, we can't prove that aa^{-1}=1 either.

On the other hand, if we start with a1=a and aa^{-1}=1 then we CAN show that 1a=1 and a^{-1}a=1 using a similar proof to those above.

I have now posted this as an alternative solution.
6th October 2009


26 Aug 2009, 12:34

Joined: 22 Apr 2010, 15:52
Posts: 43
Location: Olpe, Germany
Post Re: Exercise [13.01]
flashbang wrote:
As for the second part, I would have thought the only way really to do that would be to find a counter-example. Anyone?

Here is my proposal of a counter-example:

Consider the set M = \{ e, \alpha \} of mappings from a set S = \{A, B, C, ... \} onto itself, wherein e, \alpha are defined as follows:
e: x \mapsto A \ \forall x \in S (i.e. e maps everything ontoA )
\alpha: x \mapsto B \ \forall x \in S (i.e. \alpha maps everything ontoB )

Then e is a "right identity" in the sense that

ae = a \ \forall a \in M (proof "for all" a \in M: ee: x \mapsto A, i.e. ee = e , and \alpha e: x \mapsto B, i.e. \alpha e = \alpha)

Furthermore, the definitions e^{-1} = e and \alpha^{-1} = e provide proper "left inverses" in the sense that
a^{-1}a = e \ \forall a \in M.

However, e is not a "left identity" because
e \alpha: x \mapsto A \ \neq \  \alpha: x \mapsto B

The set M therefore has a "right identity" and "left inverses" for each element, but the "right identity" is no "left identity".


26 Apr 2010, 15:14

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [13.01]
jbeckmann, good example, but you can make it easier to understand.

Simply take a set S of (any number of) elements and define "multiplication" on the elements such that:

xy = x for all x,y in S.

Then clearly (xy)z = x(yz) = x so the associative law holds.
Any element will work as a right identity: xy=x for all x,y
so choose any particular element to be 'the' identity, 1.

then 1x = 1 for all x, so every element has the same left-inverse, and it's the identity, 1.

Clearly there is no left identity, nor right inverse, so associativity + right identity + left inverse is insufficient to guarantee these properties.


18 Jul 2010, 15:34
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