## Group axioms

Show that alo reduced axioms
• [(A)] $$\forall a,b,c\in G: (ab)c=a(bc)$$
• [(E')] $$\exists e\in G\ \forall a\in G: ae=a$$
• [(I')] $$\forall a\in G\ \exists b\in G: ab=e$$
define grup, i.e., that these provide also the "missing" rules $$ea=a$$ and $$ba=e$$.
• #### Hint

Show first from which side we can cancel common factors.

• #### Resolution

It is possible to cancel from right: $$ca=da\ \Longrightarrow\ c=ce=cab=dab=de=d$$.

(The cancelaton from left cannot be deduced in the same way, since we do not know yet tha $$c=ec$$.)

We now deduce auxiliary facts: $$e=ab=(ae)b=aabb$$, but also $$e=ee=abab$$, hence after cancelation from right: $$aab=aba$$.

Now follows: $$ea=aba=aab=ae=a$$.

Now the left cancelation could be derived: $$bc=bd \ \Longrightarrow\ c=ec=abc=abd=ed=d$$.

The last fact $$e=ab=ba$$ we get by cancelation of $$a$$ on right from the already derived fact $$aab=aba$$.