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A set $G$ has an operation $\ast$ and is closed and associative under that operation.

(a) there exists $e$ in $G$ such that $a \ast e=a$
Prove $e$ is the identity by showing $e\ast a=a$

(b) For all $a$ in $G$ there exists $b$ such that $a \ast b=e$
Prove $b$ is an inverse by showing $b \ast a=e$

For (a) I tried $e \ast e=e$, $e \ast a \ast e= a \ast e$, and $e \ast a = a$. But I'm not sure if that's allowed since I'm very new to abstract still.

For (b) I'm completely stumped. My professor gave the hint to start with $b \ast a$ and develop $e$ but I keep thinking to do it via commutative ways which I know isn't guaranteed to happen in a group. Is there a way to manipulate associativity and closure to get there?

Any help is appreciated.

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    $\begingroup$ "But I'm not sure if that's allowed" No, you may not insert things into the middle like that. You may left multiply by something to both sides of an equation or right multiply by something to both sides of an equation., but not insert. As an additional hint, you haven't yet used associativity. $\endgroup$ – JMoravitz Sep 3 '19 at 23:48
  • $\begingroup$ A set $S$ with an associative binary operation $\ast: S\times S\to S$ is called a semigroup, hence the semigroups tag. $\endgroup$ – Shaun Sep 3 '19 at 23:53
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Consider the example of $\ast$ such that $x\ast y = x$ for all $x,y$ over some set $G$ with at least two distinct elements.

Clearly, this operation is closed.

Clearly, there exists a right-identity (and in fact, everything is a right identity) $e$ such that $x\ast e = x$ for all $x$.

It should also be clear that this operation is infact associative as $(x\ast y) \ast z = x\ast z = x = x\ast y = x\ast (y\ast z)$

However, so long as $G$ has at least two distinct elements should be clear that there is no left-identity as for any proposed left identity $\ell$ and $x$ not equal to $\ell$, you would have $\ell \ast x = \ell$, not $x$.


This all shows that the conditions of closure, associativity, and existence of right identity alone are not enough to prove the existence of left-identity. For that, one would need additional assumptions.

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    $\begingroup$ perhaps to translate this into a question for the OP: what properties of a group have you been given? Which ones haven't you mentioned or used yet? $\endgroup$ – Rylee Lyman Sep 4 '19 at 0:27

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