3

This problem is from Herstein's 'Topics in Algebra'. I've thought about it a bit but haven't come up with much.

Let $G$ be a non-empty set with an associative product which also satisfies:

  1. $\exists e\in G $ such that $\forall a \in G$, $a \cdot e=a$.
  2. Given $a \in G$, $\exists y(a) \in G$, such that $y(a) \cdot a =e$.

Prove that $G$ need not be a group.

I know that $G$ is a group if any one of those multiplications is switched around, i.e, either $a \cdot y(a)=e$ or $e \cdot a=a$. But I can't quite understand why in this case $G$ is not a group. I'd be much obliged if someone can give me a good explanation of why it must be and I'd also appreciate a counter-case if nothing else.

Thanks a lot. Cheers!

Martin Sleziak
  • 56,060
sayantankhan
  • 2,447

1 Answers1

10

Consider a set with two elements $a$ and $e$. Define the product of any finite number of elements however bracketed to be equal to the first (leftmost) symbol. This ensures the product is associative.

It is straightforward (I leave it to you) to check that the two defining properties hold, and to show that this is not a group.

Mark Bennet
  • 101,769
  • Thanks a lot. That really helped. :) – sayantankhan Sep 17 '14 at 17:32
  • 1
    In fact, this works for a set with any number of elements (note that neither identity nor inverses end up being unique). – Tobias Kildetoft Sep 17 '14 at 17:50
  • 1
    @TobiasKildetoft Indeed, and a useful thing to notice. I wanted to illustrate the principle that if you don't fully understand a definition, try to build the simplest possible model and see what happens. You have built on that with "then see what you can generalise" (or if it doesn't work, what goes wrong). Here it is particularly useful to describe the product in a way which makes associativity obvious, as it saves a lot of checking. – Mark Bennet Sep 17 '14 at 19:16
  • I remember constructing this example once because someone asked about a semigroup which was not a group, but which acted transitively on itself by translation. – Tobias Kildetoft Sep 17 '14 at 19:29