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In a lecture my professor quickly went over the problem:

Let $ a $ and $ b $ be elements of a group G. If $ a^3 b = ba^3 $ and if a has order 7, show that $ ab = ba $ .

The sketch of a proof he wrote out went like this (this is what I have in my notes)

Does $ (a^3)^2 = a^6 $ commute with $ b $?

Well, $ a^6 = a^{-1} $ because $ a^7 = e $

So $ a^{-1}b = b a^{-1} $ so $ b = aba^{-1} $ so $ ba = ab $

I don't completely follow. It seems like this should be basic, but it isn't for me. By writing out $ (a^3)^2 $ it seem to imply that he could substitute $ (a^3)^2 $ for $ a^3 $ and put it into the original equation. That doesn't make sense to me because I figured $ a^3 $ is a distinct element, not a variable.

I am not sure how he is synthesizing the premises with $ a^6 = a^{-1} $ to find $ a^{-1}b = b a^{-1} $. I am hoping for some help with that.

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up vote 9 down vote accepted

$ba^3b^{-1}=a^3$ so $ba^6b^{-1}=ba^3b^{-1}ba^3b^{-1}=a^3a^3=a^6$ Thus $ba^{-1}b^{-1}=a^{-1}$; so $b$ commutes with $a^{-1}$ and so with any power of $a^{-1}$ including $a=a^{-6}$

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Another approach that doesn't involve going through the inverse: since $a^7=e$ we can write $ab = a(a^7)(a^7)b = a^{15}b$; now $$\begin{align}ab &=a^{15}b \\ &=\left(a^3\right)^5b &\text{by factoring }a^{15}\text{ into blocks of }a^3\\ &= \left(a^3\right)^4a^3b &\text{pulling out one factor of }a^3\\ &= \left(a^3\right)^4ba^3 &\text{commuting it past }b\text{ using }a^3b=ba^3\\ &= \left(a^3\right)^3b\left(a^3\right)^2 &\text{doing the same with another factor of }a^3\\ &=\ldots \\ &= b\left(a^3\right)^5 &\text{once we've pulled all the }a^3\text{s to the right}\\ &=ba^{15} &\text{packing them together again} \\ &=ba(a^7)(a^7)\\ &= ba &\text{reversing the process that got us from }a\text{ to }a^{15}\\ \end{align}$$

The same argument proves more generally that in any group, if $a$ is of order $n$, $a^p$ commutes with $b$ and $\gcd(n,p)=1$ then $a$ itself commutes with $b$.

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I feel like this deserves an upvote even though it does not answer the question directly. – peoplepower Oct 3 '12 at 3:21
It also generalises to: If $\cdot:S \rightarrow S$ is an associative binary operation and $a,b \in S$ such that $a^3b=ba^3$ and $a^8=a$, then $ab=ba$. – Douglas S. Stones Oct 3 '12 at 5:08

Given the equation $a^3b=ba^3$, multiply by $a^3$ on both sides to get $a^{-1}b=a^6b=(a^3b)a^3=ba^6=ba^{-1}$. Hence, $b$ and $a$ must commute.

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Your professor is using this fact: If $x$ commutes with $y$ (that is, if $xy=yx$), then any power of $x$ commutes with $y$. This is because you can move each $x$ factor past the $y$, one at a time: $$x^n y = x^{n-1}xy = x^{n-1}yx=x^{n-2}xyx= x^{n-2}yx^2=\ldots = yx^n.$$ Apply this in the case that $x=a^3$ and $y=b$. You're given that $a^3$ and $b$ commute, so $(a^3)^2=a^6$ and $b$ commute. But $a^7 = e$, so $a^6 = a^{-1}$. So $a^{-1}$ and $b$ commute: $$a^{-1}b = ba^{-1}$$ $$b = aba^{-1}$$ $$ba = ab.$$

A higher-level view of this is that since $3$ and $7$ have no factors in common, $a^3$ must generate the same order-$7$ subgroup as $a$. Therefore some power of $a^3$ is equal to $a$ (in fact, $(a^3)^5 = a^{15} = a$), so $a$ commutes with $b$ because it is a power of something that commutes with $b$.

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If you view $a^{-1}b=ba^{-1}$ as a group acting on itself via conjugation then $b=aba^{-1}$imlpies that $b$ is invariant under conjugation by $a\in G$ where $b$ is a fixed element of the group $G$. This means that $b^G=\{b\}$ thus $b\in Z(G)$. Let $g\in G$ and using our previous $b$ we have $gb=bg$.

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multiple by $a$ at left and right sides of both sides of the equation, we can get $ba=ab$.

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