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오리톢 [902596] · MS 2019 (수정됨) · 쪽지

2021-09-01 12:33:50
조회수 261

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게시글 주소: https://orbi.kr/00039315444

$\newcommand{\aut}{\operatorname{Aut}}\newcommand{\semi}{\rtimes}$

Q : Classify all groups of order $75$ (there are three of them).\\

Fact 1. Let $p$ be a prime and $V$ be an abelian group with the property that $pv =0$ for all $v\in V$. If $|V| = p^n$, $V$ is an $n$-dimensional vector space over a field $\Bbb F_p=\Bbb Z/p$ and $\aut(V)\simeq GL_n(\Bbb F_p)$.

Fact 2. If $V$ is an $n$-dimensional vector space over a finite field $F =\Bbb F_q$, $|GL(V)| = (q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$.

Fact 3. Assume $K$ is a cyclic group, $H$ is an arbitrary group and two homomorphisms $\varphi_1,\varphi_2:K\to\aut(H)$ s.t. $\varphi_1(K)$ and $\varphi_2(K)$ are conjugate subgroups of $\aut(H)$. If $K$ is infinite assume $\varphi_1$ and $\varphi_2$ are injective. Then $H\semi_{\varphi_1}K\simeq H\semi_{\varphi_2}K$.

A : If $G$ is an abelian group of order $75$, $G\simeq\Bbb Z/75$ or $\Bbb Z/15\times\Bbb Z/5$.

Let $G$ be a non abelian group of order $75$. By the Sylow theorem, $n_5\equiv 1\bmod 5$ and $n_5\mid 75$. Hence, $n_5 = 1$ so there is a unique Sylow $5$-subgroup denoted by $P$. Let $Q$ a Sylow $3$-subgroup. Then by the 3rd isomorphism theorem, $PQ<G$ and $|PQ| = 75$ so that $PQ =G$. Since $P\cap Q =1$, $G=P\semi Q$. Since $G$ is non abelian, induced homomorphism $\varphi:Q\to \aut(P)$ is nontrivial. Since $|Q| = 3$, $\varphi$ injects into $\aut(P)$ so $3\mid|\aut(P)|$.

If $P\simeq\Bbb Z/25$ then $|\aut(P)| = \varphi(25) = 20$ so $P$ is not cyclic.\\

Now consider $P\simeq\Bbb Z/5\times\Bbb Z/5$. By Fact 1, $\aut(P)\simeq GL_2(\Bbb F_5)$ and by Fact 2, $|GL_2(\Bbb F_5)| = 24\cdot 20 = 2^5\cdot 3\cdot 5$. Hence, we can choose $\varphi$ as a homomorphism of $P$ to Sylow $3$-subgroup of $\aut(P)$. Since $Q$ is a cyclic group and any $3$-Sylow subgroup is conjugate,  the semidirect product does not depend on $\varphi$ i.e. $P\semi Q$ is the unique nonabelian group of order $75$.\\

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