# Prove that $G$ is isomorphic to $S_4$ given $G$ is a group of order $24$ that has no element of order $6$. [duplicate]

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 $$|N_3|=6$$.

## marked as duplicate by MatheinBoulomenos, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 4 '18 at 18:32

• – user530422 Nov 4 '18 at 14:39

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.