Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

6 Answers 6

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 $a=a^{-6}$

share|improve this answer

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 $b$.

share|improve this answer
    
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

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 $b$.

share|improve this answer

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.

share|improve this answer

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 $gb=bg$.

share|improve this answer

$$a^{-1}b=a^{6}b=a^{3}a^{3}b=a^{3}ba^{3}=ba^{3}a^{3}=ba^{6}=ba^{-1}$$

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.