5

I am trying to find all groups of order $39.$ So far, I have shown that only two such groups exist ($\mathbb{Z}_{39}$ and one nonabelian group). My question is: how can I find a presentation for the nonabelian group? I know that it contains elements of order $3$ and $13$, so I need two generators. But I also need to find some relation that exists between the two generators. Is there any systematic way to go about doing this?

Kan't
  • 4,819
Analytical
  • 95

1 Answers1

9

The 13-sylow, assume generated by $x$, is normal in $G$. Let $y$ be a generator for a 3-sylow. Then by normality, we have $$yxy^{-1} = x^i$$ for some integer $i$. Hence $i^3 \equiv 1 \pmod{13}$, which implies $i\equiv 1,3,9 \pmod{13}$.

The case $i=1$ gives the cyclic group $C_3 \times C_{13}$. The groups given by $i=3,9$ are isomorphic, so a presentation is given by $$\{x,y\mid x^{13} = 1, y^3 = 1, yx= x^3y \}$$

pisco
  • 19,748
  • Why is $i^3 \equiv 1 \pmod{13}$? – Analytical Sep 27 '17 at 01:56
  • 4
    @Analytical $$yxy^{-1} = x^i \implies y^2xy^{-2} = yx^iy^{-1} = x^{i^2} \implies x=y^3xy^{-3} = yx^{i^2}y^{-1} = x^{i^3} $$ – pisco Sep 27 '17 at 01:59
  • I understand, but why do you then need it to be congruent to $1 \pmod{13}$? – Analytical Sep 27 '17 at 02:05
  • 1
    Becuase $x=x^{i^3}$, and $x$ has order 13, this implies $i^3 \equiv 1 \pmod{13}$. – pisco Sep 27 '17 at 02:07
  • I see, now it's obvious. Sorry to bother you, there's still one thing I don't understand: how do you know the cases for $i = 3, 9$ are isomorphic? – Analytical Sep 27 '17 at 02:08
  • 3
    If $yxy^{-1} = x^3$, then $y^2 x y^{-1} = x^9 $, this tells us the case $i = 9$ arises from the case $i = 3$ via replacing $y$ by $y^2$. But the subgroups generates by $y$ and $y^2$ are just the same. – pisco Sep 27 '17 at 02:11
  • The proof Is not fully clear to me. Why the $i$ coming from the 3rd Sylow thm appears as exponent of $yxy^{-1}=x^i$? By normality we should have simply $yxy^{-1}=x$ – riccardoventrella Aug 14 '22 at 15:43