# Division Laws for Groups

## Theorem

Let $G$ be a group.

Let $a, b, x \in G$.

Then:

$(1): \quad a x = b \iff x = a^{-1} b$
$(2): \quad x a = b \iff x = b a^{-1}$

## Proof

All derivations can be achieved using applications of the group axioms.

### Proof of $(1)$

 $\displaystyle a x$ $=$ $\displaystyle b$ $\displaystyle \leadsto \ \$ $\displaystyle a^{-1} a x$ $=$ $\displaystyle a^{-1} b$ $\displaystyle \leadsto \ \$ $\displaystyle e x$ $=$ $\displaystyle a^{-1} b$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle a^{-1} b$

and the converse:

 $\displaystyle x$ $=$ $\displaystyle a^{-1} b$ $\displaystyle \leadsto \ \$ $\displaystyle a x$ $=$ $\displaystyle a a^{-1} b$ $\displaystyle \leadsto \ \$ $\displaystyle a x$ $=$ $\displaystyle e b$ $\displaystyle \leadsto \ \$ $\displaystyle a x$ $=$ $\displaystyle b$

$\blacksquare$

### Proof of $(2)$

 $\displaystyle x a$ $=$ $\displaystyle b$ $\displaystyle \leadsto \ \$ $\displaystyle x a a^{-1}$ $=$ $\displaystyle b a^{-1}$ $\displaystyle \leadsto \ \$ $\displaystyle x e$ $=$ $\displaystyle b a^{-1} b$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle b a^{-1}$

and the converse:

 $\displaystyle x$ $=$ $\displaystyle b a^{-1}$ $\displaystyle \leadsto \ \$ $\displaystyle x a$ $=$ $\displaystyle b a^{-1} a$ $\displaystyle \leadsto \ \$ $\displaystyle x a$ $=$ $\displaystyle b e$ $\displaystyle \leadsto \ \$ $\displaystyle x a$ $=$ $\displaystyle b$

$\blacksquare$