In any group where $ab=b^4a$ and $b^6=e$ we have $b^3=e$

Question

Suppose $G$ is a group and $a,b$ are in $G$ and we have $ab=b^4a$ and $b^6=e$ where $e$ is the identity, then prove $b^3=e$.

I have played around with this for an hour or so by now, but the solution just keeps escaping me.

I added a more conceptual answer (that makes the proof obvious from well-know facts). — Bill Dubuque, Oct 02 '20 at 22:35

score 3 · Accepted Answer · answered Oct 02 '20 at 20:42

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We have $aba^{-1} = b^4$. Now

$$a b^3 a^{-1} = (aba^{-1})^3 = b^{12} = e$$

Hence $b^3 = a^{-1} e a = e$.

answered Oct 02 '20 at 20:42

player3236

16,413

Implicitly this is exploiting the fact that conjugation is a group isomorphism so preserves order, so then we can use well-known facts relating the order of $,b^4,$ to the order of $,b,,$ e.g. see my answer. – Bill Dubuque Oct 02 '20 at 22:35

score 2 · Answer 2 · edited Oct 02 '20 at 20:46

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Given that $ab=b^4a$, it follows that $$\begin{align} ab^3&=(ab)b^2\\ &=(b^4a)b^2\\ &=b^4(ab)b\\ &=b^4(b^4a)b\\ &=b^8(ab)\\ &=b^{12}a\\ &=ea\\ &=a. \end{align}$$

edited Oct 02 '20 at 20:46

Shaun

44,997

answered Oct 02 '20 at 20:40

Servaes

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This is so slick and instructive; the underlying theme is to notice that substitution of $b^4 a$ for $ab$ increases the power of $b$, and when that power reaches a large enough value, we can reduce it via $b^6 = e$. – heropup Oct 02 '20 at 20:45
@heropup More instructive is to exploit the fact that conjugation is a group isomorphism so preserves order, then use well-know facts relating the order of $,b^4,$ to the order of $,b,,$ e.g. see my answer. – Bill Dubuque Oct 02 '20 at 22:33
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Anyone care to explain the downvote? – Servaes Oct 02 '20 at 22:42

Bill Dubuque · Answer 3 · 2020-10-02T23:19:01.873

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Conceptually, the Key Idea is that conjugation $\,b\mapsto aba^{-1}$ is a group isomorphism so preserves group theory properties such as $\,o(b) := $ order of $b= |\langle b\rangle|.\,$ This view makes the proof obvious. OP is the special case below: $\,n,k = 6,4,\,$ so the gcd $(n,k) = (6,4)=2,\,$ so $\,j = 6/2 = 3$

Theorem $\ \ \ \color{#0a0}{ab = b^{\large k}a},\ b^{\large n} = 1\,\Rightarrow\, b^{\large j} = 1,\,$ for $\ j = n/(n,k)$

Proof $ $ Key Idea $\,\Rightarrow\, o(b) = o(\color{#0a0}{aba^{-1}}) = o(\color{#0a0}{b^k})\mid j\,$ via standard $\color{#90f}{\rm Lemma}$ relating $\,o(b^k)$ to $\,o(b)$.

edited Oct 02 '20 at 23:19

answered Oct 02 '20 at 22:31

Bill Dubuque

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Bill, fully agree, +1-ed you since this conceptual proof learns more. – Nicky Hekster Oct 02 '20 at 23:15
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@Nicky Thanks, I abstracted out the Lemma since it likely obscured the simplicity of the one-line proof. – Bill Dubuque Oct 02 '20 at 23:47

score 1 · Answer 4 · answered Oct 02 '20 at 20:41

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We have $aba^{-1}=b^4$ and hence using $b^6=e$, $$ b^4=b^{16}=(aba^{-1})^4=ab^4a^{-1} $$ So $ab^4a^{-1}=aba^{-1}$, hence $b^3=e$.

answered Oct 02 '20 at 20:41

Dietrich Burde

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I think it is shorter to do $b=a^{-1}b^4a$ thus $b^3=a^{-1}b^{12}a=a^{-1}ea=e$. – zwim Oct 02 '20 at 21:18

In any group where $ab=b^4a$ and $b^6=e$ we have $b^3=e$

4 Answers4