Prove if in a group $a$ commutes with $b$, it also commutes with $b^k$ for any integer $k$ 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.

*

*Is my proof correct?

*Can writing be improved?

*Is my use of induction appropriate?

*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.

*

*Yes, it appears so.

*I think it's perfectly fine, given what you are doing.

*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.

*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.
