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Find a non-abelian group with exactly four elements of order 5.

I'm pretty new to group theory and the best I can think of is $D_4$ has 5 elements of order 2. What's a group that satisfies the desired properties?

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  • $\begingroup$ Dihedral group of order 10. $\endgroup$ – Qiyu Wen Feb 17 '16 at 5:33
  • $\begingroup$ @QiyuWen so what are the 4 elements that have order 5? Please explain. $\endgroup$ – user3772119 Feb 17 '16 at 5:40
  • $\begingroup$ @QiyuWen In your example, if $s^2=1$ and $r^5= 1$, then what is the order of $rs$ ? $\endgroup$ – vnd Feb 17 '16 at 5:48
  • $\begingroup$ @vnd $5$ is a prime, so all non-trivial elements in $⟨r⟩$ must have order $5$. $rs^n=s^{−n}r$ for all integer $n$, so $(rs^n)^5=r^5s^N \neq e$, where $N$ is some integer. $\endgroup$ – Qiyu Wen Feb 17 '16 at 6:20
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$C_5 \times S_3$ works.

More generally, let $S$ be a finite non-abelian group of order coprime with $5$. Then $C_5 \times S$ works.

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  • $\begingroup$ What exactly are the elements of $C_5 \times S_3$ that have order 5? $\endgroup$ – user3772119 Feb 17 '16 at 6:01
  • $\begingroup$ @user3772119, $(x,1)$ with $x\ne1$. $\endgroup$ – lhf Feb 17 '16 at 6:02
  • $\begingroup$ So $(0,1), (2, 1), (3, 1), (4, 1)$? Why just those elements? And they have order 5 because $(3,1)$, for instance, has $(3,1)^5=e$? $\endgroup$ – user3772119 Feb 17 '16 at 6:09
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    $\begingroup$ @user3772119: By definition of the cartesian product, $(n,\sigma)^k = (nk~\text{mod }5,\sigma^k)$. When $k=5$, what happens to the first component? What happens to the second component for any $\sigma\neq 1$? $\endgroup$ – Eric Stucky Feb 17 '16 at 6:13
  • $\begingroup$ @EricStucky $(3,1) \rightarrow (15,1)$? $\endgroup$ – user3772119 Feb 17 '16 at 6:16
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Hint: How about starting with the group of linear transformations $\mathbb R^2 \to \mathbb R^2$ generated by the rotation $T = T_{2\pi/5}$. Of course that's Abelian. So can you think of adding another generator that won't ruin the order of $T^n$ for $n = 1,2,3,4$?

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  • $\begingroup$ what do you mean ruin the order of $T^n$? How does that group of linear transformations have 4 elements? $\endgroup$ – user3772119 Feb 17 '16 at 5:51

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