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Apologies if this question was posed already.

I need to prove that $(AB)^{k} = A^{k}B^{k}$ holds if $AB=BA$. After trying it myself, I looked at the solution and found it quite strange.

"Use induction: $(AB)^{k} = A^{k}B^{k}$ is true for $k=1$ since since $AB=BA$. Assume $(AB)^{k-1}=(BA)^{k-1}$ and prove it true for k: $$ (AB)^{k} = (AB)^{k-1} AB $$ $$ = A^{k-1}B^{k-1} AB $$ $$ = A^{k-1}A B^{k-1} B $$ $$ = A^{k} B^{k} $$"

I don't understand how this really proves anything. We had to assume $(AB)^{k-1}=(BA)^{k-1}$, how does if follow then that $(AB)^{k} = A^{k}B^{k}$ holds?

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    $\begingroup$ it should have been "assume $(AB)^{k-1} = A^{k-1}B^{k-1}$" $\endgroup$
    – Exodd
    Mar 25, 2021 at 7:47

3 Answers 3

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$$(AB)^k$$ $$=ABAB\dots AB$$ $$=AABBAB\dots AB$$ $$= AABABB\dots AB$$ $$\vdots$$ $$=AA\dots ABB\dots B$$ $$=A^kB^k$$

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@Exodd has already noted a typo in the statement of the inductive step's inductive hypothesis. Another approach to the inductive step, one that minimizes the rearrangements, is$$(AB)^k=A^kB^k\implies(AB)^{k+1}=A(BA)^kB=A(AB)^kB=AA^kB^kB=A^{k+1}B^{k+1}.$$

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  • $\begingroup$ Both answers are very helpful (+1) but I'll accept his one because it is closer to the textbook solution I asked about in my question. $\endgroup$
    – Papayapap
    Mar 25, 2021 at 8:47
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In regards to "I don't understand how this really proves anything":

I think it should be easy to see that the first case for $k=1$ holds because $(AB)^1=AB=A^1B^1$

for $k=2$, $(AB)^2 = (AB)(AB)= ABAB = A(BA)B=_{\text{because }AB=BA} A(AB)B=A^2B^2$

assuming that $(AB)^{k-1}=A^{k-1}B^{k-1}$ holds for some $k\geq 2$ uses the case $k=2$ as jumping point, because we know it is true for $k=2$.

The objective is to prove that it holds for all $k \in \mathbb{N}$, so far we only know it holds for $k=1,2$.

What about $k=3$, could we use the knowledge we have already gathered? Yes:

$(AB)^3 = (AB)(AB)(AB) = (AB)^2(AB)\color{red}{=_*}A^2B^2AB=A^2BBAB=A^2B(BA)B=A^2B(AB)B=A^2(BA)B^2=A^2(AB)B^2 = A^3B^3$

therefore we were able to prove that $(AB)^3 =A^3B^3$ by knowing that $(AB)^2=A^2B^2(\color{red}{*})$.

Instead of doing this ad infinitum and checking for $k=4,5,6,7,...$ individually, which would be impossible, tedious and boring, instead we notice a pattern:

If we know an arbitrary previous case held, like say for the case $k-1$: $(AB)^{k-1} = A^{k-1}B^{k-1}$, as in say you checked all the cases up to $k-1$ $(1,...,k-1, \text{in our case } k-1=2)$ could we show that the next case $k$ held as well?

Well, $(AB)^k=(AB)^{k-1}(AB)=A^{k-1}B^{k-1}(AB)= A^{k-1}(B^{k-1}A)B$

and because we know that $AB=BA$, given $B^{k-1}A =B^{k-2}(BA)=B^{k-2}(AB)=B^{k-3}(BA)B=B^{k-3}AB^2 = \cdots =AB^{k-1}$ we can always shift $A$ to the far left for powers of $B$.

Thus $A^{k-1}(B^{k-1}A)B=A^{k-1}(AB^{k-1})B=A^{k}B^{k}$.

In essence this is checking for $k=4,5,6,7,...$ without checking each one individually, because we just proved that given $2$ holds, which it does, $3$ holds as well, and given $3$ holds, $4$ holds as well... and so on and so on:

If $k=2$ holds, then $k=3$ holds,

If $k=3$ holds, then $k=4$ holds,

$\vdots$

If $k=k'-1$ holds, then $k=k'$ holds,

$\vdots$

So it might be weird proving if the case $k=3$ holds, then $k=4$ holds, without knowing anything about the case $k=3$ itself at first, but we know that it holds true because we proved that if the case $k=2$ held then the case $k=3$ held true, as $k=2$ was our base case, we know it's true because it was proven in the first line. All the pieces fall one after another like an infinite chain reaction...

Hope this helps!

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  • $\begingroup$ I know I shouldn't use comment to thank, but in this case really I want to thank for your effort to give an explanation of the inductive concept, this made it much clearer! $\endgroup$
    – Papayapap
    Mar 26, 2021 at 9:00
  • $\begingroup$ not a problem! glad to have helped $\endgroup$ Mar 26, 2021 at 10:15

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