I would like to know if the statement and my proof are correct, as this seems surprising to me.

Let $G$ be a group of order 24 without normal sylow subgroups.
From sylow theory, $n_2=3$, $n_3=4$. Let $G$ act by conjugation on its sylow $3$-subgroups with the corresponding homomorphism $\phi : G \to S_4$.
The action is transitive from sylow's theorems.
Further, a sylow $3$-subgroup cannot normalize any other sylow $3$ subgroup, thus any element of a sylow $3$ subgroups acts as a 3-cycle on the sylow subgroups not containing it.
Taking the square of such an element results in the inverse $3$ cycle. Thus, $\phi(G) \le S_4$ must contain all $3$-cycles in $S_4$, so it contains $A_4$. It is left to prove that it equals $S_4$.

Assume by contradiction that $\phi(G) = A_4$. then $\phi$ has a kernel of order 2. So there is a single nontrivial element in the kernel, $z$. Since $\left< z \right>$ is the kernel of $\phi$, it is normal in $G$. Consider $G/\left<z\right>$: Since $\left< z \right>$ is normal in $G$, it is contained in all three $2$-sylow subgroups of $G$. From the lattice isomorphism theorem, we can conclude that there are three $2$-sylow subgroups of $G/ \left< z \right>$.

It is simple to prove that all groups of order 12 must have some normal sylow subgroup (via counting elements). Thus, $G/ \left< z \right>$ must have a normal sylow $3$-group. Thus the images of the four $3$-sylow subgroups of $G$ under the natural projection must be equal. Thus, there are four distinct elements of $G$ whose class in $G/ \left< z \right>$ is equal, which is absurd as $ | \left< z \right> | = 2$.

Is my proof correct? Alternative proofs are welcome.

  • $\begingroup$ Please first also have a look at proofs from duplicates, e.g., here. $\endgroup$ Jan 19 '18 at 13:02
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    $\begingroup$ You can shorten this proof significantly by using the fact that $A_4$ has a normal Sylow $2$-subgroup. So if $\phi(G) = A_4$, then $G$ has a normal Sylow $2$-subgroup contrary to assumption. $\endgroup$
    – Derek Holt
    Jan 19 '18 at 13:03
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    $\begingroup$ You can even further shorten it by instead noting that the kernel of your homomorphism must be in $N_G(P)$ for all Sylow 3-subgroups $P$, and thus this kernel has order at most $2$. $\endgroup$
    – Steve D
    Jan 19 '18 at 18:35

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