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Abstract algebra

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quasar987
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Homework Statement


I am asked to show that if E is a semi-group and if

(i) there is a left identity in E
(ii) there is a left inverse to every element of E

then, E is a group.


The Attempt at a Solution


Well I can't seem to find the solution, but it's very easy if one of the two "left" above is replaced by a "right". For instance, if we replace the existence of a left inverse condition by the existence of a right inverse, then we find that the left identity is also a right identity like so:

Let a,b be in E. Then ab=a(eb)=(ae)b ==> a=ae (by multiplying by bˉ¹ from the right). So e is a right identity also. Then it follows that every right inverse is also a left inverse:

aaˉ¹=e ==>(aaˉ¹)a=ea ==>a(aˉ¹a)=a ==> (aˉ¹a)=e.

So, does anyone know for a fact that this question contains or does not contain a typo?
 

Answers and Replies

  • #2
quasar987
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Let a,b be in E. Then ab=a(eb)=(ae)b ==> a=ae (by multiplying by bˉ¹ from the right). So e is a right identity also.
No, this only means that ae=ae. :frown:
 
  • #3
Hurkyl
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Well, I can prove that all of the left inverses of e are, in fact, equal to e. So I'm making progress. :smile:
 
  • #4
Hurkyl
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Also,

x * (left inverse of x) = (left identity)

I think that the group structure follows from these facts. So I have at least as much confidence in the original problem as I do that I didn't make a mistake. (I'm not saying how much confidence that is. :wink:)
 
  • #5
Dick
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I'm having problems with this as well, but then I'm tired. I would suggest though, that if you really think it's wrong that you start trying to construct a counterexample. If you can't construct a counterexample then the effort may teach you what you need to do.
 
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  • #6
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I know that I've done this proof before, and as I recall you need to somehow use the two facts in conjuction, and remember that if y is the left inverse of x, then y also has a left inverse say z... I remember having to use this someow, but my efforts on this tonight are not going well.

Edit: With a fair amount of work I managed to prove that every left inverse is also a right inverse, and from that I think it follows a little more easily that the left identity is also a right identity.

Edit 2: It actually follows almost trivially from the fact that every left inverse is also a right inverse that the left identit is also the right identity.
 
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  • #7
Hurkyl
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Basically, I tried writing lots of expressions that could be simplified in multiple ways to derive new properties. For example, I first wondered how left inverses of the indentity (and their left inverses, etc) behaved, then I started worrying about how inverses of general elements behaved.

Incidentially, I did get started by searching for a counterexample; I decided to let 0 be the identity, 1 a left inverse of 0, 2 a left inverse of 1, and so forth, then I tried to compute how multiplication had to behave.
 
  • #8
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I think I got it. According to my calculation, the statement is true as stated. But since I'm not a mathematician and did in fact not even know the terms left-inverse and left-identity before tackling the problem (I found left-inverse in an algebra book, didn't find left-identity) I need some sanity-check:
I assumed the two conditions mean that:

[tex] \exists \, 1_L \in E : \forall a \in E : 1_L \, a = a[/tex] (i)
and
[tex] \forall \, a \in E: \, \exists \, a_L \in E: a_L \, a = 1_L [/tex] (ii).
Is this translation of the two conditions correct?
 
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  • #9
matt grime
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It is, though it is always preferable not to write things in logical forms like that since they are unnecessarily opaque.
 
  • #10
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i guess this is pretty easy:
i would go like this:

From knowing E is a semi group you have associativity!
And i know:

e*a=a , for e beeing a left identity!

and

a^-1 * a = e , a^-1 is left inverse

now i multiply a from the left and get:

a*a^(-1)*a = a

from using associativity i get:

(a*a^(-1))*a = a*(a^(-1)*a) = a (1)

hence by using both assumptions the first part of (1) implies that a*a^(-1) = e -> every left inverse is a right inverse if associativity holds!

the second part implies that while we assumes a^(-1)*a= e -> every left identity is a right identity.

QED
 
  • #11
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I don't completely understand what you said Mr. Brown. Most notably, I don't get the step
a^-1 * a = e , a^-1 is left inverse

now i multiply a from the left and get:

a*a^(-1)*a = a
which seems to imply that a*e=a.
 
  • #12
quasar987
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i guess this is pretty easy:
i would go like this:

From knowing E is a semi group you have associativity!
And i know:

e*a=a , for e beeing a left identity!

and

a^-1 * a = e , a^-1 is left inverse

now i multiply a from the left and get:

a*a^(-1)*a = a
You went a little too fast. Multiplying from the left by a gives

a*a^(-1)*a = ae

but ae is not known to be a, for e is only a left identity.
 
  • #13
quasar987
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Let me try something here. (Assuming the problem is stated correctly)

If I show that (aˉ¹)ˉ¹=a, then this will mean that e=(aˉ¹)ˉ¹(aˉ¹)=aaˉ¹, meaning left inverses are also right inverses.

Lets begin the random manipulations :)

(aˉ¹)ˉ¹(aˉ¹)=e ==>(aˉ¹)ˉ¹(aˉ¹)a=ea ==> (aˉ¹)ˉ¹e=a ==> aˉ¹(aˉ¹)ˉ¹e=aˉ¹a ==>aˉ¹(aˉ¹)ˉ¹e=e ==> aˉ¹(aˉ¹)ˉ¹=e ==> aˉ¹=((aˉ¹)ˉ¹)ˉ¹.

If every element can be seen as the left inverse of another, then I have succeeded. But is this implied? Gotta go.
 
  • #14
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[...]aˉ¹(aˉ¹)ˉ¹e=e ==> aˉ¹(aˉ¹)ˉ¹=e [...]
ae=a strikes back.
I don't think it's a good idea to label the left-inverse and left-identity [tex]a^{-1}[/tex] and e/1. That nomenclature imho cries for stupid mistakes caused by that you usually label real inverses and identities with these symbols. Might differ from person to person, but I did chose different names for exactly the reason that I screwed up too many steps otherwise.
 
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  • #15
Hurkyl
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aˉ¹(aˉ¹)ˉ¹e=e ==> aˉ¹(aˉ¹)ˉ¹=e
How'd you manage that? You have neither proven that e is a right identity, that e has a unique left inverse, or anything else I've noticed that would allow you to conclude that.
 
  • #16
quasar987
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I accepted w/o proof that

Well, I can prove that all of the left inverses of e are, in fact, equal to e.
:smile:
 
  • #17
learningphysics
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I was stuck trying to figure out this same problem recently. I looked it up online... I can't find the link right now.

But the trick is to first prove:

If x*x = x, then x = e, for any element x.

This is simple

x*x = x
x^-1 * x * x = x^-1 * x (left multiply both sides by x^-1)
e * x = e
x = e

Then you can show that every left inverse is a right inverse

x*x^-1
= x * (e * x^-1)
= x * (x^-1 * x) * x^-1 (write out e as x^-1*x)
= (x * x^-1) * (x * x^-1)

So using the previous solution we know that x * x^-1 = e, so the right inverse part is proven

So to prove e is a right identity:
x * e
= x * (x^-1 * x)
= (x * x^-1) * x
= e * x
= x

Then you can also prove that e is the unique left identity and unique right identity.
 
  • #18
quasar987
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Cheers!
 

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