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Let $G$ be a group of order $24$ that has no element of order $6$. Prove that $G$ is isomorphic to $S_4$. Claim that the number of $3-$Sylow subgroups is $4$, and consider the conjugation action of $G$ on the set of $3-$Sylow subgroups.

Any help is appreciated. I have since we claim there are four $3-$Sylow subgroups $|G:N_3|=4$ so $|N_3|=6$.

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marked as duplicate by MatheinBoulomenos, Namaste abstract-algebra Nov 4 '18 at 18:32

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Let's prove first that there are actually exactly four $3$-Sylow subgroups. By Sylow III, the number of Sylow subgroups is either $1$ or $4$. Suppose that the number of $3$-Sylow subgroups is $1$, then let $P$ be the unique (and hence normal) $3$-Sylow subgroup of $G$. Since $P$ is normal in $G$, we have a homomorphism $G \to \mathrm{Aut}(P)$ given by sending $g$ to conjugation with $g$ (as an automorphism of $P$). As $P$ is cyclic of order $3$, we have $\mathrm{Aut}(P) \cong (\Bbb Z/3\Bbb Z)^\times \cong \Bbb Z/2\Bbb Z$, this implies that there is an element $x$ of order $2$ in the kernel of the homomorphism $G \to \mathrm{Aut}(P)$, but the kernel of this homomorphism is the centralizer $C_G(P)$. If $y$ is a generator of $P$, then $x$ and $y$ commute and hence $xy$ is an element of order $6$ in $G$. This is a contradiction so $G$ has four $3$-Sylow subgroups.

Let $P_1, P_2, P_3, P_4$ be the $3$-Sylow subgroups of $G$, then we can define a homomorphism $G \to S_4$ by letting $G$ act on $\{P_1,P_2,P_3,P_4\}$ by conjugation. Let $K$ be the kernel of this homomorphism.

We have $$K = \{g \in G\mid \forall i \in \{1,2,3,4\}: gP_ig^{-1} = P_i\} = \bigcap_{i=1}^4 N_G(P_i)$$

We have that $P_i$ is a $3$-Sylow subgroup of $N_G(P_i)$. As $P_i \cap P_j = \{1\}$ for $i \neq j$ (since the $P_i$ have order $3$ which implies that they are simple), we get that $K$ doesn't contain any element of order $3$. As $|N_G(P_i)|=6$, the only remaining possibilities are $|K|=2$ and $|K|=1$.

Suppose that $|K|=2$, then $K$ is a normal subgroup (as it is the kernel of a homomorphism) of order $2$ and hence central (see this question), so if we take a generator $x$ of $K$ and a generator $y$ of $P_1$, then $x$ and $y$ commute and so $xy$ has order $6$, contradicting our assumption.

Thus $|K|=1$, so the homomorphism $G \to S_4$ is injective and thus an isomorphism, as $G$ and $S_4$ have the same order.

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