위수 12인 군과 위수 24인 군은 단순군(simple group)이 아님을 알 수 있습니다.
문제 1
Prove that every group of order 12 has a normal subgroup of order 3 or 4.
Proof. Let $G$ be a group of order $12=2^2\cdot 3$, then Sylow 2-subgroups of $G$ have order $2^2$, and Sylow 3-subgroups of $G$ have order 3. Let $n_2, n_3$ be the number of Sylow $n_2, n_3$-subgroups of $G$. By the Sylow theorem, there are integers $k_2,k_3\ge 0$ such that $n_2=2k_2+1\mid 3$ and $n_3=3k_3+1\mid 4$. Thus $n_2=1\text{ or }3$ and $n_3=1\text{ or }4$. If $n_3=1$, then the Sylow 3-subgroup is normal, so $G$ has a normal subgroup of order 3. Now suppose that $n_3=4$. Sylow 3-subgroups intersect trivially, so $G$ has at least 8 elements of order 3 and at most 3 elements that are not of order 3. Thus there is at most one Sylow 2-subgroups of $G$. Thus $G$ has a normal subgroup of order 4.
문제 2
Prove that every group of order 24 has a normal subgroup of order 4 or 8.
Proof. Let $G$ be a group of order $12=2^3\cdot 3$, then Sylow 2-subgroups of $G$ have order $2^3$. Let $n_2$ be the number of Sylow 2-subgroups, then $n_2=1$ or $n_2=3$. If $n_2=1$, then the Sylow 2-subgroup is normal, so $G$ has a normal subgroup of order 8. Now suppose $n_2=3$, and let $H$ and $K$ be distinct Sylow 2-subgroups. Let $N=N_G(H\cap K)$ be the normalizer of $H\cap K$. Then
Since $H\cap K$ is a subgroup of $H$, the order of $H\cap K$ is a divisor of 8. Thus $H\cap K$ has order 4. Since
$H\cap K$ is normal in both $H$ and $K$, so $H,K\subset N$ and so $HK\subset N$. Since $N$ is a subgroup of $G$, the order of $N$ divides 24. Also,
Thus the order of $N$ must be 24 and so $N=G$. Therefore, $H\cap K$ is normal in $G$, and $G$ has a normal subgroup of order 4.