Restrictions of Groups of Small Order (6.2.18-20)

Dummit and Foote Abstract Algebra, section 6.2, exercises 18-20:
18. Prove $|G|=36⇒n_2 = 1 \lor n_3 = 1$.
19. Prove $|G|=12~\land \nexists H(H ≤ G \land |H| = 6)⇒G≅A_4$.
20. Prove $|G|=24~\land \nexists g(g∈G \land |g| = 6)⇒G≅S_4$.

Proof: (18) Sylow analysis reveals:$$n_2∈\{3,9\}~~~~~n_3=4$$Let $G$ act by left multiplication on the four cosets of $N_G(P_3)=P_3$; the kernel $N$ of this action is the largest normal subgroup of $G$ contained in $P_3$. Since $P_3 \not \trianglelefteq G$ and $N=1$ allows $G$ isomorphic passage into $S_4$ of order 24, we must have $|N|=3$. We have $|G/N|=12$, so there is either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup of $G/N$. Assuming the latter, we have $K/N \trianglelefteq G/N$ is of order 3, so that its preimage $K$ is of order 9 and normal in $G$, a contradiction. Therefore let $K/N \trianglelefteq G/N$ be of order 4 so that $K \trianglelefteq G$ is of order 12. It contains some Sylow 2-subgroup of $G$, and by its normality it contains all of them, so $n_2=3$. Let $K$ act by left multiplication on the three cosets of $P_3$ it contains; once again, we must have a normal subgroup $T$ of order 2 in $K$. This can only be the 2-core of $K$, which is characteristic in $K$, and now normal in $G$. We have $TN$ is a normal subgroup of order 6, so that $P_3(TN)$ is a subgroup; since $P_3 \not ≤ TN$ and $P_3∩TN=1$ implies $P_3(TN)$ is a subgroup of order 54, we must have $|P_3∩TN|=3$ and $P_3(TN)$ is a subgroup of order 18. This subgroup is normal by its index, contains a Sylow 3-subgroup and thus contains all of them, and yet Sylow purports $n_3=1$.$~\square$

(19) Since $G \not ≅ A_4$ by assumption, we have $n_3=1$. Take $x$ of order 2 and we have $\langle~x~\rangle P_3$ is a subgroup of order 6.$~\square$

(20) We have:$$n_2∈\{1,3\}~~~~~n_3∈\{1,4\}$$If $n_3=4$, then letting $G$ act on the cosets of $P_3$ allows isomorphic passage into $S_4$ so that by order $G≅S_4$. In addition, assume $n_2=1$; then $G=P_2P_3 ≅ P_2 \times P_3$. If $Z(P_2)=P_2$, then $Z(G) ≅ Z(P_2) \times Z(P_3) = P_2 \times P_3$, so that $G$ is abelian and the elements $x$ and $y$ of orders 2 and 3 respectively combine for a product $xy$ of order 6. If $|Z(P_2)|=2$ (the only other option for $Z(P_2)$), then $Z(P_2)~\text{char}~P_2 \trianglelefteq G$ so that $Z(P_2) \trianglelefteq G$ and now $Z(P_2)P_3 ≅ Z_6$. Ultimately, $n_2=3$. Since $3 \not ≡ 1 \mod{2^2}$ we have $P_2∩Q_2$ of order 4; by exercise 13, we must have $N_G(P_2∩Q_2)=G$ so that $P_2∩Q_2 \trianglelefteq G$, and by 4.5.37 $P_2∩Q_2$ is contained in every Sylow 2-subgroup, so that the sum of the distinct elements of order 2, 4, and 8 can be explicitly calculated; there is the intersection point of order 4 minus the identity, and the three Sylow 3-subgroups that each provide 4 unique elements (lest the intersection between two distinct Sylow 2-subgroups be greater than 4, an impossibility), yielding 15 such elements. Comparing this with the following chart of possible orders of elements that can be determined thus far:we invariably end up with only 18 elements of $G$, a contradiction.$~\square$