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

Question

This is problem from I.N Herstein Page 35 Q3 .How should i start doing this ?Hints ? Thanks

Answer 1

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.

Answer 2

I am not sure what is the dot operations and if it's the same as group binary operation, but I will assume that it is, anyway it's group so there could be only one operation. $$ (a \cdot b)^2 = a^2 \cdot b^2 $$

$$ (a \cdot b) (a \cdot b) = a^2 \cdot b^2 $$ $$ a \cdot b \cdot a \cdot b = a \cdot a \cdot b \cdot b $$

First multiply this element by $b^{-1}$ from right and then with $a^{-1}$ from left.

$$ b \cdot a = a \cdot b $$

So it's abeliean group, if my assumtions were good.

Answer 3

$(ab)^2=abab=a^2b^2 \rightarrow ab=ba $