Let $G$ be a set with associative binary operation and a unit.

Question

Let $G$ be a set with associative binary operation and a unit. Assume that for every $ g \in G$ there exists $ x \in G$ with $xg = 1$. Prove that $gx = 1$ is a consequence.

That above is the question, and i think i have the answer however I need clarification on the unit part of the question.

So by saying there is a unit in the set, does this mean that the neutral element is 1, and that the statement is saying that for $ x \in G$ there exists a left inverse $x$ such that $xg = 1$. And conversely we can say that for every $x$ there exists a left inverse say $g'$ such that $g'x=1$. And then by associativity i have proved the consequence.

Is this the right way to go and the simplest?

score 1 · Answer 1 · 2013-05-12T13:14:57.490

1

Denote $x$ by $x_g$. then for each $g$: $$gx_g=1gx_g=x_{x_g}x_ggx_g=x_{x_g}1x_g=x_{x_g}x_g=1$$

edited May 12 '13 at 13:14

answered May 12 '13 at 12:48

axiom of choice is used when you index. – May 12 '13 at 13:17
sorry i have never tried it like that could you explain it further? – tedg May 12 '13 at 13:28
for each $g$ there's some $x_g$ with $x_gg=1$. if $g'=x_g$ then $x_{g'}g'=1$ that is $x_{x_g}x_g=1$. – May 12 '13 at 13:31

Let $G$ be a set with associative binary operation and a unit.

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