Thread: [SOLVED] Abstract algebra semigroup question

Let $\displaystyle \left({S, \circ}\right)$ be a semigroup.

Let $\displaystyle u \in S$ be such that:

$\displaystyle \forall a \in S: \exists x, y \in S: u \circ x = a = y \circ u$

Prove that $\displaystyle \left({S, \circ}\right)$ has an identity element.

i think, you can work on with this..

since $\displaystyle u \in S$, $\displaystyle \exists \, x,y \, \in S$ such that $\displaystyle u \circ x = u = y \circ u$

hi everybody,
actually this questions has got nothing to do with the above quetsion, which is above all my understandung, by the way..
but im new here so i kinda dont no how to deal with this forum example where to post my questions and where to go to may be help someone.. can any one help?

Originally Posted by Zara

hi everybody,
actually this questions has got nothing to do with the above quetsion, which is above all my understandung, by the way..
but im new here so i kinda dont no how to deal with this forum example where to post my questions and where to go to may be help someone.. can any one help?

if you don't know where to put your question, go to Urgent Homeworks folder and click New Thread to post your things.. the mods will put them in their proper places.. however, if you know that it belongs to calculus, algebra whatsoever, then find the folder and post your questions there.

many helpful friends will read it and help you do it..

EDIT: i think you already know what to do..

thanks a lot!!!!!!!!!!!!!!!!!!!!!!
and how can i help?like, how do i come to those questions that havent been answered?

sorry mods for being off-topic..

Originally Posted by Zara

thanks a lot!!!!!!!!!!!!!!!!!!!!!!
and how can i help?like, how do i come to those questions that havent been answered?

just read them, and if you can do them, reply with your partial, i mean partial, solutions.

Thanks Kalagota - I've got that far but haven't got further.

I think I can establish that if there's an identity element then it needs to be $\displaystyle u$.

I think that if I can then establish that both $\displaystyle x$ and $\displaystyle y$ need to be equal to $\displaystyle a$ then the job is done.

I haven't been able to rule out the possibility that $\displaystyle x$ and $\displaystyle y$ need not be equal to $\displaystyle a$.

And I can get no further.

Originally Posted by Matt Westwood

Let $\displaystyle \left({S, \circ}\right)$ be a semigroup.

Let $\displaystyle u \in S$ be such that:

$\displaystyle \forall a \in S: \exists x, y \in S: u \circ x = a = y \circ u$

Prove that $\displaystyle \left({S, \circ}\right)$ has an identity element.

If $\displaystyle \forall\, a \in S,\ \exists\,x, y \in S: u \circ x = a = y \circ u$, then since $\displaystyle u\in S$, we have that $\displaystyle \exists\,v,w\in S$ such that $\displaystyle u\circ v=u=w\circ u$.

Let a be any element in S. Then there are $\displaystyle x,y\in S$ with $\displaystyle u\circ x=a=y\circ u$.

$\displaystyle a=u\circ x\ \Rightarrow\ w\circ a=w\circ u\circ x=u\circ x=a$

$\displaystyle a=y\circ u \Rightarrow\ a\circ v = y\circ u\circ v = y\circ u = a$

Hence $\displaystyle w\circ a = a\;\ldots\;\fbox{1}$ and $\displaystyle a\circ v = a\;\ldots\;\fbox{2}$ for any $\displaystyle a\in S$. In particular, letting $\displaystyle a=v$ in $\displaystyle \fbox{1}$ gives $\displaystyle w\circ v=v$ and letting $\displaystyle a=w$ in $\displaystyle \fbox{2}$ gives $\displaystyle w\circ v = w$.

Thus $\displaystyle v=w\circ v=w$ is the identity element in S.

That was so elegant. Much indebted.