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Let $G$ be a set with a binary operation *, associating to each pair of elements $x$ and $y$ of $G$ a third element $x*y$ of $G$. Suppose that the following properties are satisfied:

  1. $(x*y)*z = x*(y*z)$ for all elements $x$, $y$, and $z$ of $G$ (the Associative Law);
  2. there exists an element $e$ of $G$ such that $e*x = x$ for all elements $x$ of $G$;
  3. for each element $x$ of $G$ there exists an element $x'$ of $G$ satisfying $x'*x = e$.

How to prove

i. $x*e=x$ and

ii. $x*x'=e$,

so that $G$ is a group?

velut luna
  • 9,961

2 Answers2

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$x'(xe)=(x'x)e=ee=e=x'x$, $x''x'xe=x''x'x$, $xe=x$ takes care of i.

Gerry Myerson
  • 179,216
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This is Nicolas Bourbaki's Algebra I, Chapter 1, §4, Ex.2 (a).

As for every $x$ there exists $x'$, such that $x'x=e$, so $x'xy=ey$, that is, $x'(xy)=y$, hence left translation by $x$ is injective.

Next, $\;\;ee=e \Rightarrow (x'x)e=x'x\Rightarrow x'(xe)=x'x \Rightarrow xe=x\;$. So $e$ is an identity element. Also, $x'xx'=ex'=x'e\Rightarrow xx'=e$.

Hence $x$ is invertible. Thus $G$ is a group.

nature1729
  • 551