Group of Order 72 with Trivial Center

Exercise. $|G|=72$, $Z(G)=1$. Characterize $G$.

In fact, we obtain all possibilities of $G$ by magma.

Let $S$ be the set of minimal normal subgroups of $G$. Then $S\subseteq \{{\mathbb{Z}}_2,{\mathbb{Z}}_2^2,{\mathbb{Z}}_2^3,{\mathbb{Z}}_3,{\mathbb{Z}}_3^2\}$. Note that for any $N\in S$, $N$ can be seen as an irreducible $G/C_G(N)$-module. Otherwise, if $N$ is reducible, then there exists $M<N$ such that $M^G=M$ and so $M\in S$, $N\notin S$.

If ${\mathbb{Z}}_2\in S$, then $Z(G)\supseteq {\mathbb{Z}}_2$, which contradicts with $Z(G)=1$. Thus ${\mathbb{Z}}_2\notin S$.

If ${\mathbb{Z}}_2^2\in S$, then by NC lemma $G/C_G({\mathbb{Z}}_2^2)\lesssim \mathrm{Aut}({\mathbb{Z}}_2^2)=\mathrm{GL}(2,2)$. Thus $|C_G({\mathbb{Z}}_2^2)|⩾12$ and $3\mid |C_G({\mathbb{Z}}_2^2)|$. If $9\mid |C_G({\mathbb{Z}}_2^2)|$, then ${\mathbb{Z}}_2^2$ is reducible, contradiction. Hence, ${\mathbb{Z}}_3$ is a characteristic subgroup of $C_G({\mathbb{Z}}_2^2)$. Since $C_G({\mathbb{Z}}_2^2)⊲G$, we have that ${\mathbb{Z}}_3⊲G$ and so $G$ has a normal subgroup of ${\mathbb{Z}}_2^2\times {\mathbb{Z}}_3$. Then $G=({\mathbb{Z}}_2^2\times {\mathbb{Z}}_3).Q$ where $Q\in \{C_6,D_6\}$.

If ${\mathbb{Z}}_2^3\in S$, define $A:=G/C_G({\mathbb{Z}}_2^3)\lesssim \mathrm{GL}(3,2)$. Then $Z(A)$ induces a set of isomorphisms of the module ${\mathbb{Z}}_2^3$.
By Schur’s lemma, $Z(A)\cong {\mathbb{Z}}_1$ or ${\mathbb{Z}}_7$. Then $|A|\mid 72$ yields that $Z(A)=1$. By magma, there does not exist an irreducible $H⩽\mathrm{GL}(3,2)$ such that $Z(H)=1$ and $|H|\mid 72$. Therefore, ${\mathbb{Z}}_2^3\notin S$.

If ${\mathbb{Z}}_3\in S$, then $|C_G({\mathbb{Z}}_3)|=36$ and $C_G({\mathbb{Z}}_3)$ is a normal subgroup of $G$.

If ${\mathbb{Z}}_3^2\in S$, then $G={\mathbb{Z}}_3^2.Q$ with $|Q|=8$. Let $P$ be a Sylow $2$-subgroup of $G$. Then $P\cap {\mathbb{Z}}_3^2=1$ and so $G={\mathbb{Z}}_3^2:P$, where $P\in \{{\mathbb{Z}}_8,Q_8,D_8\}$.

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D:=SmallGroupDatabase();
A:=[s:s in SmallGroups(D, 72)|#Center(s) eq 1];
#A;
[GroupName(a):a in A];

for a in A do
    Order(MinimalNormalSubgroup(a));
    GroupName(MinimalNormalSubgroup(a));
    "------";
end for;

G:=GL(3,2);
L:=Subgroups(G);
L:=[l: l in L|#Center(l`subgroup) eq 1];
L:=[l: l in L|(72 div Order(l`subgroup))*Order(l`subgroup) eq 72];

for l in L do
    IsIrreducible(l`subgroup);
end for