# Prove there is no element of order 6 in a simple group of order 168

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?

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Why does $n_2=21$, mean there is no element of order 6? – user23238 May 6 '13 at 22:33

## 1 Answer

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...

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Why does an element of order 6 belong to exactly one normalizer? – nullUser Oct 11 '11 at 18:51
@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 6:13
@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 6:50