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I was told that if $|ab|=5$ then $a^5 b^5 = e$ but only if we know the group is abelian. I would like a possible example of a nonabelian group where this equality holds.

4 Answers 4

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Take $S_5$ the symmetric group on 5 elements and any 5-cycle $a$. Then $a^2$ is also a 5-cycle, while $a^5a^5=ee=e$.

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    I can see that it made more sense to read the request as for the given equality to hold everywhere rather than just at one pair $a,b$.2012-10-21
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There are non-Abelian $5$-groups of exponent $5$. For example, let $G = \{ \left( \begin{array}{cl} 1 & x & y \\ 0 & 1& z \\0&0&1 \end{array}\right): x,y,z \in \mathbb{Z}/5\mathbb{Z} \}$. In such a group, we have $(ab)^{5} =1 \neq ab$ for all $a,b \in G^{\#},$ such that $b \neq a^{-1}.$

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    Yes indeed, you are correct Hagen. I really meant to say that that element genuionely has order $5$, except in the stated cases. I will edit to make clearer.2012-10-21
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I suppose it's (somewhat vacuously) true for all non-abelian groups $G$ with orders indivisible by $5$, so let's skip those cases.

Result: There is a unique smallest non-abelian group of order divisible by $5$ for which $|ab|=5$ implies $a^5 b^5=\mathrm{id}$, namely $\mathbb{Z}_5 \times S_3$

In a sense, this is a cop-out, since we take an abelian group $\mathbb{Z}_5$, which must satisfy the property (since it's abelian), and we append $S_3$, which makes it non-abelian, but does not affect the property, since $|S_3|=6$ is coprime to $5$.

I used both Mace4 and GAP, to find there are no examples of orders 5,10,15,20, and 25. I also checked, of the three non-abelian groups of order $30$, the only one that satisfies the desired property is $\mathbb{Z}_5 \times S_3$. We can also prove the property holds as follows:

If $(u,x), (v,y) \in \mathbb{Z}_5 \times S_3$ and $|(u,x)(v,y)|=5$, then, since $|S_3|=6$ is coprime to $5$, we must have $xy=\mathrm{id}$.

Hence \begin{align} (u,x)^5 (v,y)^5 &= (u^5 v^5,x^5 x^{-5}) & \text{since } y=x^{-1} \\ &= ((uv)^5,(xx^{-1})^5) & \text{since } u,v \in \mathbb{Z}_5, \text{ and } x \text{ and } x^{-1} \text{ commute} \\ &= ((u,x)(v,x^{-1}))^5 \\ &= \mathrm{id} & \text{since } |(u,x)(v,y)|=5. \\ \end{align}

Actually, the same proof works to generate infinitely many examples:

Result: If $G$ is an abelian group and $H$ is a non-abelian group of order $|H| \not\equiv 0 \pmod 5$, then $G \times H$ is a non-abelian group for which $|ab|=5$ implies $a^5 b^5=\mathrm{id}$.

The smallest examples this result does not cover have order $125$ (found by GAP), which are:

  • $\mathbb{Z}_{25} \ltimes \mathbb{Z}_5$,
  • $(\mathbb{Z}_5 \times \mathbb{Z}_5) \ltimes \mathbb{Z}_5$; which is isomorphic to the example given by Geoff Robinson.
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    Very nice guys I really wish that I could give everyone best answer2012-10-21
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In $S_{10}$ consider two disjoint 5-cycles $a$, $b$. Then $a$, $b$, $ab$ all have order 5. (Admittedly, $\langle a, b\rangle is an abelian subgroup here.)