# If $A,B$ are square matrices of the same order, given, $A^4=I,A^2=B^2,AB=BA^3$, where $I$ is the identity matrix, show $A^3B=BA$ and $A^2B=BA^2$.

The question is:

$$A$$ and $$B$$ are square matrices of the same order.

Given that, $$A^4=I, A^2=B^2, AB=BA^3$$, where $$I$$ is the identity matrix.

Show that $$A^3B=BA$$ and $$A^2B=BA^2$$.

My workings:

1). For showing $$A^3B=BA$$

Using $$AB=BA^3$$, multiplying on right By$$A$$ $$\implies ABA=BA^4$$

$$\therefore ABA=B$$, since $$A^4=I$$

Multiplying on left by $$A^3 \implies A^3ABA=A^3B$$

$$\therefore A^4BA=A^3B \implies BA=A^3B$$

$$\therefore A^3B=BA$$ ... Result!

2). For showing $$A^2B=BA^2$$

$$A^2B=A(AB)$$, and $$A(AB)=A(BA^3)$$, from given $$AB=BA^3$$,

and, $$A(BA^3)=A(BA)A^2$$

$$\therefore A^2B=A(BA)A^2 ... (1)$$

From the given $$AB=BA^3$$, multiplying on the right by $$A$$ gives,

$$ABA=BA^4 \implies ABA=B$$, since $$A^4=I$$

$$\therefore (1)$$ becomes,

$$A^2B=BA^2$$ ... Result!

• You could prove this using group theory; presentations, in particular. Commented Jun 17, 2023 at 8:06

The second equation is much easier to prove: $$A^2B=B^2B=BB^2=BA^2$$.

For the first one, notice that since $$A^4=I$$, $$A$$ is invertible. $$AA^3B=A^4B=IB=BA^3A=ABA$$, left multiplied by $$A^{-1}$$, we got $$A^3B=BA$$.

• Thanks for positive input, and another insight!
– Nik
Commented Jun 17, 2023 at 10:13

According to GAP, the group given by

$$P:=\langle A,B\mid A^4=I, A^2=B^2, AB=BA^3\rangle$$

is the quaternion group $$Q_8$$:

gap> F:=FreeGroup(2);
<free group on the generators [ f1, f2 ]>
gap> rels:=[ (F.1)^4, (F.1)^2*(F.2)^(-2), (F.1)*(F.2)*(F.1)^(-3)*(F.2)^(-1) ];
[ f1^4, f1^2*f2^-2, f1*f2*f1^-3*f2^-1 ]
gap> G:=F/rels;
<fp group on the generators [ f1, f2 ]>
gap> Size(G);
8
gap> StructureDescription(G);
"Q8"


Furthermore, it can be computed using GAP whether those identities hold:

gap> (G.1)^3*(G.2)=(G.2)*(G.1);
true
gap> (G.1)^2*(G.2)=(G.2)*(G.1)^2;
true


This can be seen more directly by mapping $$A\mapsto i$$ and $$B\mapsto j$$ in the quaternions.

• I know! You could always upvote it to compensate, @user1176409 ;) Commented Jun 17, 2023 at 8:43
• Yeah I couldn't understand the downvote! I'm only wanting to contribute POSITIVELY to this site, not end up feeling disheartened by a NEGATIVE experience!
– Nik
Commented Jun 17, 2023 at 10:09