Fun problem sets from Pinter: h3-8, p43. h7 requires "stroke of genius". The rest are elementary.

h3

If xax = e, then (xa)^2n = a^n

xax = e
=> xaxa = a
 (xa)^2 = a
 (xa)^2n = a^n
 QED.

h4

If a^3 = e, then a has square root.

Proof:

NTS a^(1/2) = some j => a = j^2
a^3 = e
=> a^2 = a^-1
=> a = (a^-1)^2 = j^2
QED

h5.txt

If a^2 = e, then a has cube root.

Proof:

NTS a^(1/3) = some j => a = j^3
a^2 = e
=> a = a^-1 ( multiply by a^-1 on both sides )
=> e = (a^-1)^2 ( multiply by a^-1 on both sides )
=> a = (a)(a^-1)^2 ( multiply by a on both sides)
=> a = (a^-1)^3 (since a = a^-1 )
QED

h6.txt

If a^-1 has cube root, so does a.

Proof:

a^-1 = j^3. NTS a = k^3
a = (j^3)^-1 ( invert both sides )
  = (j(jj))^-1 (write j^3 as j(jj) )
  = (jj)^-1 j^-1  (since (ab)^-1 = b^-1a^-1) 
  = (j^-1)(j^-1)(j^-1) = k^3 (since (ab)^-1 = b^-1a^-1)
QED
 

h7.txt

If x^2ax = a^-1 , show a has a cube root.

Proof:

   x^2ax = a^-1
=> ax^2ax = e  ( multiply by a on both sides )
=> (ax)(xax) = e
=> (xax)^2 = x ( multiply by x on both sides )
=> (xax)^3 = xax^2  (multiply by xax on both sides )

Now lets simplify rhs
Since  x^2ax = a^-1
x^2 = a^-1x^-1a^-1

So rhs xax^2 = xa(a^-1x^-1a^-1) = a^-1
Voila!

Therefore (xax)^3 = a^-1
=> a^-1 has a cube root => a has a cube root (per h6)
QED.

h8

If xax = b then ab has square root

Proof:

NTS ab = z^2

xax = b
axax = ab
QED