A proposition which gives us an equivalence to the group definition.

Question

Let be $G$ a non-empty set and $*$ one law of composition inner. Suposse that $*$ is:

1) $\forall a, b, c \in G$ $a*(b*c)=(a*b)*c$;

2) $\exists e \in G$; $a*e=a=e*a$, $\forall a \in G$

3) $\forall a \in G $ $ \exists a' \in G$ ; $a'*a=e$. (opposite only the left )

Show that $G$ is group.

What is the question? Do you want to show that in a group left and right inverses are equivalent? — Eoin, Mar 07 '15 at 03:57
Related (more general) question: Right identity and Right inverse implies a group — Martin Sleziak, Mar 07 '15 at 06:19
And also these posts: Any Set with Associativity, Left Identity, Left Inverse is a Group AND — Martin Sleziak, Mar 07 '15 at 06:54

coffeemath · Accepted Answer · 2015-03-07T04:40:09.797

4

Given $a$ pick $a'$ with $a'*a=e.$ Then using this $a'$ choose $a''$ for which $a''*a'=e.$ Now note that $$a*a'=e*(a*a')=(a''*a')*(a*a')=\\ a''*(a'*a)*a'=a''*e*a'=a''*a'=e.$$

edited Mar 07 '15 at 04:40

answered Mar 07 '15 at 04:34

coffeemath

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A proposition which gives us an equivalence to the group definition.

1 Answers1