3

A group of order $75=3\cdot 5^2$ has a normal $5$-sylow subgroup $H_5$ and a sylow $3$-sylow subgroup $H_3$. As $H_3\cap H_5 = \{1\}$, we must have, by the recognition theorem, $$G=H_5H_3\simeq H_5\rtimes_{\varphi} H_3$$ with the map $$\varphi:H_3\to\operatorname{Aut}(H_5)(x\mapsto\varphi_x:y\mapsto xyx^{-1}).$$

Now $H_3\simeq \Bbb Z_3$. The semidirect product is non-abelian if $\varphi$ is not trivial, and it's easy to see if $H_5\simeq \Bbb Z_{25}$, $\varphi$ must be trivial. So it remains to examine the case $H_5\simeq \Bbb Z_5\times \Bbb Z_5$ with $\operatorname{Aut}(H_5) \simeq \operatorname{GL}_2(\Bbb Z_5).$

So we need all homomorphisms $$\varphi:\Bbb Z_3\to\operatorname{GL}_2(\Bbb Z_5),$$ in other words we need to find all order $3$ elements in $\operatorname{GL}_2(\Bbb Z_5)$. Definitely there will be more than one order $3$ elements in $\operatorname{GL}_2(\Bbb Z_5)$. So how do I prove that such non-trivial $\varphi$ is actually unique?

user108580
  • 767
  • 4
    Here are some quick hints. A Sylow $3$-subgroup of ${\rm GL}_2(5)$ has order $3$. All such subgroups are conjugate, and mapping onto conjugate subgroups results in isomorphic semidirect products, so we can fix ${\rm im}(\varphi)$ to be a specific subgroup of ${\rm GL}_2(3)$. That reduces to just two possible $\varphi$, but since we can choose either nonzero element as generator of $Z_3$ , they also give rise to isomorphic semidirect products. – Derek Holt Oct 05 '23 at 15:31
  • See also @KCd's first comment here – Anne Bauval Oct 05 '23 at 15:38
  • @DerekHolt “mapping onto conjugate subgroups results in isomorphic semidirect products“ that solved my problem. Thank you. – user108580 Oct 05 '23 at 15:42
  • 1
    Good question! You can also use some linear algebra to show all order-3 elements of $\mathrm{GL}_2(\Bbb Z_5)$ are conjugate (so in fact you don't even have to worry about such elements that lie in the same cyclic subgroup). The way you do that is by considering what possible RCF such an element can have. Observe that its minimal polynomial must have degree $2$ and must divide $X^3 - 1 = (X - 1)(X^2 + X + 1)$, and that $X^2 + X + 1$ is irreducible mod 5. So there is only one choice - the companion matrix of $X^2 + X + 1$! – Izaak van Dongen Oct 05 '23 at 16:04

0 Answers0