Proof: if $AB=BA$, prove that $(AB)^2 = A^2B^2$

Question

I got a little stuck with this proof. It is given that $AB=BA$.

Proof: $(AB)^2 = A^2 \cdot B^2$

I have been thinking of several ways to solve it. I got to this point:

$(AB)^2 = AB\cdot AB = BA\cdot AB = B\cdot A^2\cdot B$

But I don't know how to proceed. I think I am missing some general rule.

Can somebody help me?

How about $(AB)^2 = (AB)(AB) = A(BA)B = A(AB)B = (AA)(BB) = A^2 B^2$? Are the steps clear, or would you like me to explain further? — Kenny Wong, Jun 26 '17 at 19:11
You swapped the first $AB$ to $BA$. What if you swapped the middle $BA$ to $AB$? — Matthew Leingang, Jun 26 '17 at 19:11

score 5 · Answer 1 · answered Jun 26 '17 at 19:11

5

Just $$(AB)^2=ABAB=AABB=A^2B^2$$

answered Jun 26 '17 at 19:11

Michael Rozenberg

score 4 · Accepted Answer · answered Jun 26 '17 at 19:12

4

$$(AB)^2=(AB)(AB)=A(BA)B=A(AB)B=(AA)(BB)=A^2B^2$$ After the third equal sign we introduced $AB=BA$.

answered Jun 26 '17 at 19:12

Dave

If so it should be $(AB)^2=(AB)(AB)=((AB)A)B=(A(BA))B=(A(AB))B=((AA)B)B=(AA)(BB)=A^2B^2$. – Michael Rozenberg Jun 27 '17 at 01:45

score 2 · Answer 3 · answered Jun 26 '17 at 19:13

2

You can do:

$$ABAB$$ $AABB$ (by reversing the inner two) $$A^2B^2$$

answered Jun 26 '17 at 19:13

D.R.

3 Answers3