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오리톢 [902596] · MS 2019 · 쪽지

2021-10-19 19:28:28
조회수 371

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게시글 주소: https://orbi.kr/00040114858

(a) By Sylow theorem, $n_7\equiv 1\bmod 7$ and $n_7\mid 168$ so $n_7 = 8$. Hence, there are total $(7-1)\cdot 8 = 48$ many elements of order $7$ in $G$.\\

(b) Let $H = N_G(P_7)$. Since $n_7 = 8$, $[G:H] = 8$ so that $|H| = 21$ where $P_7$ is Sylow $7$-subgroup. Note that by Sylow theorem, $n_3 = 4,7,28$. If $n_3 = 4$ then $G$ contains a subgroup of index $4$, say $K$. If we act $G$ on the collection of cosets $G/K$, then there is an induced homomorphism $G\to S_4$. Since $G$ is simple, the kernel of this map is trivial so that $G<S_4$ which is a contradiction. Note that by definition, the only subgroup of order $7$ in $H$ is $P_7$ since any subgroup of order $7$ in $H$ is Sylow $7$-subgroup of $G$. Now let $a\in P_7$ of order $7$ and suppose $C_H(a) = H$. Then by definition, any element of $H$ commutes with any elements of $P_7$. In particular, $P_3<H$ for some Sylow $3$-subgroup of $H$ (hence for $G$) so that $P_7$ elementwise commutes with $P_3$. Hence, $P_7<N_G(P_3)$. In particular, $7$ does not divide $[G:N_G(P_3)] = n_3$. But $n_3 = 7$ or $28$ which is a contradiction. Hence, $C_H(a)\neq H$. Now consider $a^H$ which is a conjugacy class of $a\in P_7$ in $H$, $|a^H| = [H:C_H(a)] = 3$ as $P_7<C_H(a)\neq H$. If we let $b\in P_7$ with $b\notin a^H$, then $|b^H| = 3$ so that $P_7 = a^H\cup b^H\cup \{e\}$. In particular, there are exactly two conjugacy classes of elements of order $7$ in group $H$. If $g_1\in a^H$ and $g_2\in b^H$ such that for some $g\in G$, $gg_1g^{-1}  = g_2$ then $g\in H$ which is a contradiction. Hence elements of order $7$ from different conjugacy classes of group $H$ are not conjugate in $G$. Since any two Sylow $7$-subgroups are conjugate in $G$, this shows that the number of conjugacy classes of elements of order $7$ in $G$ is $2$.\\


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