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Let $G $ be a group and $ a, b\in G $ such that $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$.

How can we show that $a=b$$\,$?

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    @Deroty: (1) Are you sure the claim is true? (2) Where does the problem come from? – 2017-02-18
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    The group operation is multiply? Sorry I'm slow. – 2017-02-18
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    So in other words, you're not sure if it's true. If that's the case, you shouldn't stated the problem as "How can we show ...?". Instead, so as not to be misleading, you should state it as "Must it be true that ...?" – 2017-02-18
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    @marshal craft: Yes, multiplication. – 2017-02-18
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    Assume $b^2a=b^3a$ and $a^2b=a^3b$ then $b=1$ same for $a$. – 2017-02-18
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    @marshal craft: Multiplication need not be commutative. If $a,b$ commute, the problem is trivial. – 2017-02-18
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    @quasi I had a professor who asked us to prove a false statement once. Not sure if it was by mistake but he said he wanted to see who can prove the impossible. So its certainly on us to not assume questions are nice. A student had to show wheils first attempt at Fermat's last theorem was false. – 2017-02-18
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    @quasi, yes good for you. – 2017-02-18
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    It is true. In fact more is true: the relations imply that $a=b=1$. – 2017-02-18
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    This is a duplicate of http://math.stackexchange.com/questions/66573 – 2017-02-18
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    I'e just had a look at it, answer is not easy to follow... In fact I do not manage to reproduce User641 equations. – 2017-02-18

1 Answers 1

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$\bbox[5px,border:2px solid]{\begin{array}{c}b^2a=ab^3\\a^2b=ba^3\end{array}}\ $ these relations are equivalent to OP's since elements in a group are invertible.

If $G$ is commutative or if $a,b$ commute

We have $ab=ba$.

$b^3=a^{-1}b^2a=a^{-1}ab^2=b^2\Rightarrow b=1$

$a^3=b^{-1}a^2b=b^{-1}ba^2=a^2\Rightarrow a=1$

We get $a=b=1$.

G any group, but either $a$ or $b$ is of order $\le$ 4

For instance $a=1$

$b^3=a^{-1}b^2a=b^2\Rightarrow b=1$


For instance $a^2=1$

$a=a^3=b^{-1}a^2b=b^{-1}b=1$ and we conclude like previously


For instance $a^3=1$

$b=ba^3=a^2b\Rightarrow a^2=1$ and we conclude like previously


For instance $a^4=1$

$(ba)a^2=ba^3=a^2b=a^2ba^4=a^2(ba^3)a=a^2(a^2b)a=a^4(ba)=ba\Rightarrow a^2=1$ and we conclude like previously


Everything being symetrical in $a,b$, the same conclusion arises if we make the hypothesis on $b$.

So here also we get $a=b=1$

G any group, but either $a$ or $b$ order is finite

Let assume a^n=1

$a^2(ba)=(a^2b)a=(ba^3)a=(ba)a^3\Rightarrow \bbox[5px,border:2px solid]{\forall k\in\mathbb N,\ a^{2k}(ba)=(ba)a^{3k}}$

If n even then n=2p

$a^{2p}(ba)=a^n(ba)=(ba)=(ba)a^{3p}=(ba)a^na^{p}=(ba)a^{p}\Rightarrow a^{p}=1$ and since $p 

If n odd then n=2p+1

$a^{2p}(ba)=a^{n-1}(ba)=a^{-1}(ba)=(ba)a^{3p}=(ba)a^na^{p-1}=(ba)a^{p-1}\Rightarrow (ba)=(ab)a^p$

$b=(ab)a^{p-1}$


$(ab)b^2=ab^3=b^2a=(ab)a^{p-1}\,(ab)a^{p-1}\ a=(ab)a^{p-2}(a^2b)a^p=(ab)a^{p-2}(ba^3)a^p$

$b^2=a^{p-2}ba^{p+3}=a^{p-4}(a^2b)a^{p+3}=a^{p-4}(ba^3)a^{p+3}=a^{p-4}ba^{p+6}$

$\bbox[5px,border:2px solid]{\forall k\in\mathbb N,\ b^2=a^{p-2k}\,b\,a^{p+3k}}$

If p even

For $p=2k$ we have $b^2=ba^{5k}\Rightarrow b=a^{5k}$ then $a,b$ commute and we can conclude.

If p odd

For $p=2k+1$ we have $b^2=aba^{5k+1}\Rightarrow (ab)b^2=ab^3=b^2a=aba^{5k+2}=(ab)a^{5k+2}$

$b^2=a^{5k+2}\Rightarrow ab^3=b^2a=a^{5k+3}\Rightarrow b^3=a^{5k+2}=b^2\Rightarrow b=1$ and we can also conclude.


Everything being symetrical in $a,b$ the same conclusion arises if we make the hypothesis $b^n=1$.

Thus we have proved it in all cases either directly if $n$ odd, either by recurrence if $n$ is even.

G infinite non-commutative group, order of $a,b$ infinite

I have no idea... :-(

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    "root of unity" is not the right terminology. Presumably you meant "square root of unity", but more precise would be "element of order 2". – 2017-02-18
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    @quasi you are right, I edit my post. thanks. – 2017-02-18
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    "we assume that $a$ and $b$ [are] invertible" - aren't they invertible by definition, since they are in a group? – 2017-02-18
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    @Ankoganit Thx, right, I'm tired. I delete first paragraph, it's useless, I was thinking about counter examples and confused myself,lol. – 2017-02-18
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    @zwim: On the line after "if $n$ odd then $n=2p+1$", I follow everything except the final claim $(ba)=(ab)(a^p)$. How do you get that? – 2017-02-18
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    $a^{-1}(ba)=(ba)a^{p-1}$ multiply by $a$ left and regroup on right. – 2017-02-18
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    It's the "regroup on the right" that I don't see – 2017-02-19