If G is a group such that $(a.b)^{2}=a^{2}.b^{2}$ for all a and b,Then show that G is abelian This is problem from I.N Herstein Page 35 Q3 .How should i start doing this ?Hints ? Thanks
In groups, you can "cancel", as long as you do it from the same side.

Explicitly:

$ac = bc \implies a = b$, and $ab = ac \implies b = c$

Proof:

$ac = bc \implies (ac)c^{-1} = (bc)c^{-1} \implies a(cc^{-1})= b(cc^{-1}) \implies ae = be \implies a = b$

$ac = bc \implies a^{-1}(ac) = a^{-1}(bc) \implies (a^{-1}a)b = (a^{-1}a)c \implies eb = ec \implies b = c$.

Now you are given that $(ab)^2 = a^2b^2$, that is:

$a(ab)b = a(ab)b$. Use the above.
$(ab)^2=abab=a^2b^2 \rightarrow ab=ba $