퀄

(i) First we show the existence of $H$. From Sylow theorem, $n_3= 1$ or $7$. If $n_3 = 1$ then Sylow $3$-subgroup $P$ is normal in $G$. If $L$ is size $2$ subgroup of $G$, then $PL<G$ with order $6$. If $n_3 = 7$ then $[G:N_G(P)] = 7$. Hence, $N_G(N)<G$ is a subgroup of order $6$.\\

Now we show any two such subgroups are conjugate in $G$.\\

Let $H<G$ be an arbitrary subgroup of order $6$. Note that by Sylow theorem, $n_7=1$. Hence, if we denote $P$ be the Sylow $7$-subgroup of $G$, then $P\triangleleft G$.\\

If $G$ is abelian then it's a direct product of Sylow subgroups. Let $K,L$ be Sylow $2$ and $3$ subgroups respectively. Then $G\simeq K\times L\times P$. Since any subgroup of $G$ is of the form $K'\times L'\times P'$ where $K'<K,L'<L$ and $P'<P$, there is only one subgroup of order $6$. Since $G$ is abelian, this proves the statement.\\

Suppose $G$ is not abelian. If $H$ is abelian then $G\simeq H\times P$ which is a contradiction as $P,H$ are both abelian. Hence $H$ is not abelian. Now, $H<N_G(H)<G$ and as $N_G(H)\neq G$, $|N_G(H)| = 6$ i.e., $N_G(H) =H$. Hence, the number of conjugacy class of $H$ is $[G:N_G(H)] = 7$. Now, let $X$ be the collection of all order $6$ subgroup of $G$. Let $P$ act on $X$ by conjugation. Then by Orbit-stabilizer theorem, $|P\cdot H| = [P:P_H]$. But note that $P_H =\{p\in P\mid pHp^{-1} = H\}$ so $P_H<N_G(H)$. Since $P_H<P$ and $\gcd(7,6)=1$, we conclude $P_H =\{e\}$. Hence, $|P\cdot H| =7$. Since we know the number of conjugacy class of $H$ in $G$ is $7$, the action $P$ on $H$ is free. Recall that as $P$ is a $7$-group, we have

$$|X^P|\equiv |X|\bmod 7$$

Since $X^P =\{H\in X\mid pHp^{-1} = H,\forall p\in P\}$ and the action is free, $|X^P| =0$. Hence, $|X| = 7n$ for some $n\in\Bbb N$. Note that if there are $m$ distinct subgroup of order $6$, then there are at least $6m-3m+3 = 3m+3$ many distinct elements. Since $|G| = 42$, $3m+3\leq 42$ if and only if $m\leq 13$. In particular, $7n\leq 13$. Hence, $n =1$ i.e. $|X| = 7$ which shows that any subgroup of order $6$ are conjugate.\\

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(ii) Since $N$ is Sylow $7$-subgroup, it's normal in $G$. Let $H$ be any size $6$ subgroup of $G$. Then $G = HN$ and $H\cap N =1$ so $G = N\rtimes H$.

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