Is there a non-abelian group where $|ab|=5$ such that $a^5 b^5 = e$ 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.
 A: 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 $b \neq a^{-1}.$
A: 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.

A: In $S_{10}$ consider two disjoint 5-cycles $a$, $b$. Then $a$, $b$, $ab$ all have order 5. (Admittedly, $\langle a, b\rangle<S_{10}$ is an abelian subgroup here.)
A: Take $S_5$ the symmetric group on 5 elements and any 5-cycle $a$. Then $a^2$ is also a 5-cycle, while $a^5a^5=ee=e$.
