It is an exercise in Kargapolov to prove that
a set *G* with a binary operation · is a *group*,
if

- the operation is associative;
- the operation guarantees left and right quotients;
i.e. for each pair of elements
*a*,*b*in*G*there are*G*-elements*x*,*y*—called respectively*left*and*right quotients*of*b*by*a*—such that*ax = b*,*ya = b*.

(**R**^{+}, · ) ≅ (**R**, +)
via *x* ↦ ln *x*,
but (by M. A. Armstrong's Groups and Symmetry (1988), pp.34–5)
(**Q**, +) ≇ (**Q**^{+}, ×),
for were φ : **Q** → **Q**^{+}
an isomorphism,
then for some *x* in **Q** we'd have
2 = φ(*x*) = φ(*x*/2 + *x*/2) = φ(*x*/2)^{2},
which is impossible because φ(*x*/2) is rational
and √2 is not.

Three exercises (1.2.2–1.2.4):

- If
*a*then |^{n}= e*a*| divides*n*. - If
*ab = ba*then |*ab*| = |*a*|·|*b*|. - If |
*a*| =*m*and |*b*| =*n*then*G*contains an element—not always*ab*—whose order is lcm(*m*,*n*).

**Proposition.** If *a*, *b* ∈ *G* then |*ab*| = |*ba*|.

**Proof.** If |*ba*| < |*ab*| then
1
= (*ab*)⋯(*ab*)

(|*ab*| times)
= *a*(*ba*)⋯(*ba*)*b*

(⩾ |*ba*| times)
= (*ab*)⋯,
contradicting the minimality of the order of |*ab*|. ▮

**Alternate proof.** One may note that
1
= (*ab*)^{|ab|} ⇒ *b*^{−1}*a*^{−1}
= (*ab*)^{|ab|−1},
so that
(*ba*)^{|ab|}
= *b*(*ab*)^{|ab|−1}*a*
= *bb*^{−1}*a*^{−1}*a*
= 1,
whence |*ba*| ⩽ |*ab*|. Reversing the roles of *a* and *b* establishes the result. ▮