# Proof

• Because of the existence of inverse elements it is $$e=e\ast e^{-1}$$.
• On the other hand, because $e$ is the unique neutral element, we have $$e^{-1}=e\ast e^{-1}$$.
• Comparing both equations leads to $$e^{-1}=e$$.

• The element $$x^{-1}$$ has the inverse elements $$(x^{-1})^{-1}$$ and $$x$$, respectively.
• Both must be identical, since inverse element are unique.

• $$a^{-1}\ast b$$ is the solution of $$a\ast x=b$$, since $$a\ast (a^{-1}\ast b)=(a\ast a^{-1})\ast b=e\ast b=b$$ for all $$a,b\in G$$.
• Moreover, it is the only solution. For if an element $y\in G$ solves $$a\ast y=b$$, it follows
$\begin{array}{rcl} a\ast y&=&b\\ a^{-1}\ast (a\ast y)&=&a^{-1}\ast b\\ (a^{-1}\ast a)\ast y&=&a^{-1}\ast b\\ e\ast y&=&a^{-1}\ast b\\ y&=&a^{-1}\ast b \end{array}$

Exercise, in analogy to 3)

• If $a\ast x=a\ast y$ then for all $a\in G$: $\begin{array}{rcl} a\ast x&=&a\ast y\\ a^{-1}\ast (a\ast x)&=&a^{-1}\ast (a\ast y)\\ (a^{-1}\ast a)\ast x&=&(a^{-1}\ast a)\ast x\\ e\ast x&=&e\ast y\\ x&=&y \end{array}$
• A similar cancellation property can be concluded for the equation $x\ast a=y\ast a$, from which it follows $x=y$ for all $a\in G.$

The element $$(x\ast y)$$ has the inverse element $$(x\ast y)^{-1}$$. It follows for all $$x,y\in G$$: $\begin{array}{rcl} (x\ast y)\ast(x\ast y)^{-1}&=&e\\ y^{-1}\ast x^{-1}\ast (x\ast y)\ast(x\ast y)^{-1}&=&y^{-1}\ast x^{-1}\ast e\\ y^{-1}\ast (x^{-1}\ast x)\ast y\ast(x\ast y)^{-1}&=&y^{-1}\ast x^{-1}\\ y^{-1}\ast e\ast y\ast(x\ast y)^{-1}&=&y^{-1}\ast x^{-1}\\ y^{-1}\ast y\ast(x\ast y)^{-1}&=&y^{-1}\ast x^{-1}\\ (y^{-1}\ast y)\ast(x\ast y)^{-1}&=&y^{-1}\ast x^{-1}\\ e\ast(x\ast y)^{-1}&=&y^{-1}\ast x^{-1}\\ (x\ast y)^{-1}&=&y^{-1}\ast x^{-1}\\ \end{array}$

• "$\ast$" is associative by definition of a group (a group is a monoid, which is a semigroup, which is associative)