TIFR-2012-Maths-B: 17

Question

True/False Question:

If the equation

$$xyz=1$$

holds in a group $G$, does it follow that

$$yxz=1$$.

Answer 1

Since g is a group so every element will have a unique inverse element which may not be the case for a normal algebraic structure.
xyz=1
$x^{-1}$ xyz $z^{-1}$=$x^{-1}$$z^{-1}$
→ $y$ = $x^{-1}$$z^{-1}$
→ yx=$x^{-1}$$z^{-1}$x
→ yxz=$x^{-1}$$z^{-1}$xz

Since the group is not abelian we cannot cancel $x^{-1}$x and $z^{-1}$z so the given conclusion is false

Answer 2

Assuming 1, is identity element, then

$xyz = 1 \implies yz = x^{-1} \implies (yz)^{-1} = x \implies z^{-1}y^{-1} = x$

Suppose $yxz = 1$, is True, then $ xz = y^{-1} \implies x = y^{-1}z^{-1}$

Let $y^{-1} = a \text{ and } z^{-1} = b$, our original statement now becomes

if $ba = x$, then $ab = x$, which is clearly not true for all groups.

Comments

It will be $z^{-1}$$y^{-1}$ = x in the first line