Counterexample for $(a*b)^2=a^2*b^2$ I'm stuck on finding a counterexample for the following problem,
If $G$ is a group with operation * and $a$ and $b$ belong in $G$, then $(a*b)^2=a^2*b^2$.
I think it's false because the law of exponents only works if the group is Abelian and they never stated whether this group was Abelian. My line of thought is that I shouldn't use multiplication or addition as my operation since they are commutative but I don't know where to go from there.
 A: When constructing such examples, always try to start small. Take the smallest nonabelian group: $S_3$.
$(12), (23) \in S_3$ and $(12)(23) = (123)$.
So $[(12)(23)]^2 = (132)$. but $(12)^2 = (23)^2 = 1$, and since $(132) \not= 1$, you have the desired example.
A: An example from the group $(\mathbb H\setminus\{0\},\cdot)$:
$(ij)^2=k^2=-1$ 
but
$i^2j^2=(-1)^2=1$.
A: Consider $D_6$, the dihedral group of order 6 the smallest non-abelian group
$
\begin{array}{c|cccccc}
 * & e & a & b & c & d & f \\
\hline
 e & e & a & b & c & d & f \\
 a & a & e & d & f & b & c \\
 b & b & f & e & d & c & a \\
 c & c & d & f & e & a & b \\
 d & d & c & a & b & f & e \\
 f & f & b & c & a & e & d \\
\end{array}
$
$(a*b)^2=f^2=d$
$a^2*b^2=e*e=e$
and $d\ne e$
so the property is invalid
Hope this helps
A: Take any non-Abelian group $(G,*)$. Now take any two elements $a$ and $b$ such that $a*b\neq b*a$. Then $a*a*b*b\neq a*b*a*b$. In other words, $a^2*b^2\neq(a*b)^2$.
For instance, $G$ can be the set of all bijections from $\mathbb Q$ into itself and $*$ can be the composition. Take $a(x)=2x$ and $b(x)=1-x$.
A: $(a*b)^2 = a^2 * b^2$ if and only if $a*b*a*b=a*a*b*b$, which is if and only if $b*a=a*b$, or if and only if $a$ and $b$ commute.
Find any non-abelian group, and you will have an example.
A: We know that $2 \times 2$ invertible matrices with real entries form a group under matrix multiplication.
Just try two invertible matrices and you'll probably see that their product of squares is different from the square of their product.
