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Let $G$ be a simple group of order 168. Let $n_p$ be the number of Sylow $p$ subgroups in $G$.

I have already shown: $n_7 = 8$, $n_3 = 28$, $n_2 \in \left\{7, 21 \right\}$

Need to show: $n_2 = 21$ (showing there is no element of order 6 In $G$ will suffice)

Attempt so far: If $P$ is a Sylow-2 subgroup of $G$, $|P| = 8$. Assume for contradiction that $n_2 = 7$. Then the normalizer $N(P)$ has order 24. Let $k_p$ be the number of Sylow-$p$ subgroups in $N(P)$. Then $k_3 \in \left\{1,4 \right\}$ and $k_2 \in \left\{1,3 \right\}$. Then I showed $k_3 = 4, k_2 = 1$. Counting argument shows there is an element of order 6 in $N(P)$, and thus in $G$ too.

I don't know how to proceed from here.

I am told that there cannot be an element of order 6 in $G$, but I don't know how to prove it, if someone could help me prove this I would very much appreciate it. Can someone help me?

nullUser
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1 Answers1

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If there is an element of order 6, then that centralizes the Sylow $3$-subgroup $P_3$ generated by its square. You have already shown that $|N(P_3)|=168/n_3=6$. Therefore the normalizer of any Sylow $3$-subgroup would have to be cyclic of order 6, and an element of order 6 belongs to exactly one such normalizer. Thus your group would have $56=2\cdot n_3$ elements of order $3$, $56=2\cdot n_3$ elements of order $6$, $48=6\cdot n_7$ elements of order $7$, and therefore only eight other elements. Those eight would have to form a Sylow $2$-subgroup, and that would be unique, so...

Jyrki Lahtonen
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  • Why does an element of order 6 belong to exactly one normalizer? – nullUser Oct 11 '11 at 18:51
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    @nullUser: A $3$-Sylow is normal in its normalizer, so it is the unique $3$-Sylow subgroup in its normalizer. If $x$ has order $6$, and belongs to the normalizer of both $P$ and $Q$, then since $x^2$ lies in $P$ (as it must lie in some $3$-Sylow subgroup of $N(P)$, and $P$ is the only one), and it also lies in $Q$ (same argument, but with $N(Q)$), then $x^2\in P\cap Q$. So $P\cap Q\neq {e}$, which means $P=Q$, since they have order $3$. – Arturo Magidin Oct 11 '11 at 18:55
  • "Therefore the normalizer of any Sylow 3-subgroup would have to be cyclic of order 6." How did we conclude this? – nullUser Oct 14 '11 at 06:13
  • @nullUser: Given that there is an element of order 6, call it $x$, we know that everything in $H=\langle x \rangle \simeq C_6$ commutes with everything in the Sylow 3-subgroup $P_3=\langle x^2 \rangle$. Therefore $H\subseteq N(P_3)$. On the other hand the OP had already concluded that $|N(P_3)|=6$. Hence we can conclude that $H=N(P_3)$. If $P'$ is any other Sylow 3-subgroup of $G$, then we know that there exists $g\in G$ such that $P'=gP_3g^{-1}$. It is easy to see that then also $N(P')=gN(P_3)g^{-1}=gHg^{-1}$ is generated by an element $gxg^{-1}$ of order 6. – Jyrki Lahtonen Oct 14 '11 at 06:50
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    Is there a different approach to show that a finite group of order 168 has no elements of order 168? I saw a question where you had to show that there were no elements of order 6 before you calculated $n_3$. In this question, the question already established that $G$ is isomorphic to a subgroup of $A_8$ by the conjugation action on $Syl_7(G)$ and then say that elements of order two have no fixed points. (If it did, then it was possible to show that there will be elements of order 14 which doesn't exist in $A_8$.) – daruma Apr 24 '19 at 07:43