zyz's quiz

Exercise. Assume that $G$ acts on $\Omega$ faithfully. If $A⩽G$ is abelian and $A$ is transitive on $\Omega$, then $C_G(A)=A$.

\begin{proof} By Frattini’s argument, $G=A{G}_{\omega }$. If there exists $g\in C_G(A)\setminus A$, then $g\in G_{\omega }$ for some $\omega \in \Omega$. Since ${\omega }^{ag}={\omega }^{ga}={\omega }^a$ for all $a\in A$, we have that $g$ acts on $G$ trivially, contradiction. \end{proof}

Exercise. Show that finite subgroups of the multiplicative group of a field are cyclic.

\begin{proof} See here or Finite Group of Field. \end{proof}

• Exercise. Let $G$ be a finite group, and let $H<G$ be a finite group. Assume that $[G:H]=n$. Then there exists $\{x_1,\cdots ,x_n\}\in G$ such that $\{H{x}_i:i\in \{1,\cdots ,n\}\}$ is the set of right cosets, and $\{x_{i}H:i\in \{1,\cdots ,n\}\}$ is the set of left cosets.

\begin{proof}

\end{proof}

Exercise. Let $G=⟨a,b⟩$. If $a^3=b^3=(ab{)}^2=1$, then $G\cong A_4$.

\begin{proof} Compute $|G|$ directly.

• length $0$: $\{1\}$
• length $1$: $\{a^i,b^i:i\in 1,2\}$
• length $2$: $\{a^i{b}^j:i,j=1,2\}\cup \{ba,b{a}^2\}$
• length $3$: $\{a{b}^{2}a\}$ (note that $a{b}^{2}a=a^{2}b{a}^2=b{a}^{2}b=b^{2}a{b}^2$)
• length $⩾4$: $\varnothing$

So $G$ is of order $12$. Define $\phi :G\to A_4$ where $a\mapsto (123)$ and $b\mapsto (234)$. Since $\phi$ is surjective and $|G|=|A_4|$, then $\phi$ is a isomorphism and so $G\cong A_4$. \end{proof}

Remark. Similarly, we can prove: Let $G=⟨a,b⟩$. If $a^4=b^2=(ab{)}^3=1$, then $G\cong S_4$.

Exercise. Let $G=⟨a,b⟩$. If $a^5=b^2=(ab{)}^3=1$, then $G\cong A_5$.

\begin{proof} \end{proof}

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