Prove if in a group $a$ commutes with $b$, it also commutes with $b^k$ for any integer $k$ (based on Gallian's Algebra text).

This site has similar questions for individual integers; I'd like to prove it for all integers. My proof is below.

  1. Is my proof correct?
  2. Can writing be improved?
  3. Is my use of induction appropriate?
  4. Is there a more direct proof than using induction?

Proof: Since $ab=ba$, $aba^{-1}=b$ and similarly $a^{-1}ba=b$.

We now show that for any integer $k$, $a^kba^{-k}=b$. If $k>0$, we have $a^kba^{-k} = a^{k-1}(aba^{-1})a^{-(k-1)} = a^{k-1}ba^{-(k-1)}$. By induction, we get $a^kba^{-k}=b$. A similar induction can be used for $k<0$. And for $k=0$, $a^kba^{-k}=b$ is trivial.

Consequently, $a^kb=ba^k$, QED.

  • $\begingroup$ It seems correct to me. $\endgroup$
    – GReyes
    Sep 14 at 22:46
  • $\begingroup$ You've proved that $b$ commutes with $a^k$. Apart from that (harmless) switch of $a$ and $b$, your proof is fine, except that it doesn't deal with the case of negative $k$ (which is quite easy using $a^{-k} = (a^{-1})^k$ and what you already have). $\endgroup$
    – Rob Arthan
    Sep 14 at 22:56
  • $\begingroup$ As a minor (whimsical aside): when I read Gallian Algebra, I wondered whether this was about some new field of algebra" that I had never heard of. If you'd written "Gallian's book Algebra" or just "Gallian Algebra", I wouldn't have been taken aback. $\ddot{\smile}$. $\endgroup$
    – Rob Arthan
    Sep 14 at 23:00
  • $\begingroup$ @RobArthan I intended to cover negative $k$ in "similarly $a^{−1}ba=b$.... A similar induction can be used for $k<0$." If that is not adequate, please explain why not. $\endgroup$ Sep 14 at 23:36
  • $\begingroup$ Your "similarly ..." completion of the solution is fine. $\endgroup$
    – Rob Arthan
    Sep 14 at 23:37
  1. Yes, it appears so.
  2. I think it's perfectly fine, given what you are doing.
  3. I think there's a much more general statement that's equally easy to prove: the collection of all elements that commute with $x$ forms a subgroup of a group.
  4. This is 3. Suppose that $x$ and $y$ commute with $a$. Then $xy$ and $x^{-1}$ commute with $a$. To see this, we simply calculate: $$ axy=xay=xya,$$ and for inverses we just multiply $ax=xa$ on left and right by $x^{-1}$ to obtain $x^{-1}a=ax^{-1}$.

Now that the collection of all elements that commute with $a$ is a subgroup, the result is obvious.

If you want to just prove your result, you can use the inverse trick to show that $a$ commutes with $b^{-1}$, and then you have proved the negative powers by proving the positive powers.

  • $\begingroup$ Your explanation hides the induction needed to prove that if $H$ is a subgroup of $G$ and $h \in H$, then $h^k \in H$ for any $k$. $\endgroup$
    – Rob Arthan
    Sep 14 at 22:52
  • $\begingroup$ @RobArthan I would hope that that fact, that groups are closed under taking powers of an element, is considered more basic than the question. $\endgroup$ Sep 14 at 22:58
  • $\begingroup$ The question specifically asks whether there is a more direct method than using induction. Hiding the induction inside an assumed lemma is a less direct, although a much more structured approach. If people ask for a direct proof and you respond with a less direct but better structured proof, then good for you but you should explain what you have done. $\endgroup$
    – Rob Arthan
    Sep 14 at 23:07
  • $\begingroup$ @RobArthan If you consider the fact that powers of elements belong to a group to be a secret induction, and therefore all subsequent statements are induction-tainted, then I challenge you to provide even a definition of $a^k$ that does not use induction. Hence, it would be impossible to write down the question without using induction, hence there can be no induction-free proof. But that's silly. My proof does not use induction, beyond the statement that $a^k$ is an element of any subgroup containing $a$. That is part of the definition of being a group. $\endgroup$ Sep 14 at 23:13
  • $\begingroup$ I am sorry. My concerns are more with the question than your answer. Our last two comments taken together will hopefully explain to the OP that there can be no "direct proof that does not use induction" but that a good proof (like yours) can be given that consigns the induction to a basic lemma that is required for any non-trivial reasoning about the notation $a^k$. $\endgroup$
    – Rob Arthan
    Sep 14 at 23:23

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