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An associative * on a set G with unique right identity and left inverse proof enough for it to be a group ?Also would a right identity with a unique left inverse be a group as well then with the same logic? I don't need the proof , more so an explanation of why or why not the second case would be true or not?

Alzebrian
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  • when you say unique left identity what does that mean? – Asinomás Sep 13 '20 at 01:26
  • @JorgeFernándezHidalgo A semigroup can have multiple left identities. For example, consider the right projection operation $a\odot b=b$. This is associative, so gives a semigroup (once we fix some set to work over), but every $a$ is a left identity for this operation. – Noah Schweber Sep 13 '20 at 01:27
  • I have rephrased my query. – Alzebrian Sep 13 '20 at 01:31
  • you changed it to something else didn't you? – Asinomás Sep 13 '20 at 01:36
  • Yea I was confused about the two possibilities we would get , I wrote the opposite of what I had a query on , if that's what you're asking – Alzebrian Sep 13 '20 at 01:42
  • "I don't need the proof , more so an explanation of why or why not the second case would be true or not?" – Alzebrian Sep 13 '20 at 01:59
  • In your second question, if the right identity is not unique, how do you define "left inverse"? – bof Sep 13 '20 at 04:49
  • oh okay so would that be why the second statement is false? – Alzebrian Sep 13 '20 at 17:22
  • Nice question @Alzebrian ! As you've suggested, I know that you are definitely going to need the "unique" bit here, as left inverses together with right identity doesn't imply group (my go to example is x*y = y). But I'm not sure what the best way is to think of unique inverses. Does a unique left inverse imply a unique left identity? If you have lots of left identities, you will get lots of left inverses targeting each identity, I think? – user214962 Sep 13 '20 at 21:36

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The second situation does not parse very well as written, as “inverse” refers to identity and if you have potentially multiple identities, then you don’t know what “unique inverse” refers to.

You could sharpen it to something like the following:

Let $G$ be a semigroup such that there exists $e\in G$ such that $xe=x$ for all $x$, and such that for each $f\in G$ that is a right identity and each $x\in G$ there exists a unique $y\in G$ such that $yx=f$. Is $G$ a group?

The answer is “no”. Let $G$ be any set with more than one element and define the operation on $G$ by $xy=x$ for all $x,y\in G$. Then every element of $G$ acts as a right identity, and given any $f,x\in G$, there is a unique $y\in G$ such that $yx=f$ (namely, $y=f$). However, $G$ is not a group.

As to the original question, it has been asked before.

Arturo Magidin
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