if $ab=e$ Then $ba=e$ and $b=a^{-1}$

Question

Let $G$ by a group and let $e$ be it's identity element, and let $a\in G$ Prove:

1.If $aa=a$ then $a=e$

2.If $b\in G$ such that $ab=e$ then $ba=e$ and $b=a^{-1}$

Proof:

1. $$aa=a$$

$G$ is a group then there is $a^{-1}\in G$ such that $a^{-1}a=aa^{-1}=e$

$$a^{-1}aa=a^{-1}a\iff a^{-1}aa=e$$

Now, $G$ is a group then operation is associative, for every $a,b,c\in G: (ab)c=a(bc)$

$$\begin{align} a^{-1}(aa)=e&\iff (a^{-1}a)a=e\\ &\iff ea=e\\ &\iff a=e. \end{align}$$

2.

$$ab=e$$

$$\begin{align} a^{-1}ab=a^{-1}e &\iff a^{-1}(ab)=a^{-1}\\ &\iff (a^{-1}a)b=a^{-1} \\ &\iff eb=a^{-1}\\ &\iff b=a^{-1}, \end{align}$$

$$ab=e,$$

$$\begin{align} abab=ab&\iff a^{-1}abab=a^{-1}ab\\ &\iff bab=b\\ &\iff bab^{-1}=bb^{-1}\\ &\iff ba=a. \end{align}$$

Is it correct?

Answer 1

Shortly:

$$ab=e\iff a^{-1}ab=a^{-1}e\iff b=a^{-1}$$

and

$$ab=e\iff baba=bea\iff (ba)(ba)=ba\iff ba=e,$$ using the first result with $ba$.

Answer 2

1. If we denote by $u$ an inverse of $a$ (we still don't know whether this is unique, so I find "$a^{-1}$" somehow misleading at this stage), and recall that group operation is a map, we get:

$$aa=a \Rightarrow (aa)u=au=e \Rightarrow a(au)=e \Rightarrow ae=a=e$$

2. $ab=e$ is precisely the definition of inverse of $a$, so $b=a^{-1}$ is just a (smart indeed, in hindsight) notation for this element once you have proved that it is unique.