Frattini Subgroups and the Burnside Basis Theorem — Examples

Every $p$th power in $P$ is a power of $g^p$, and conversely $g^p$ is a $p$th power, so [F1] gives $P^p=⟨g^p⟩$. The group is abelian, so $P^{\mathrm{\prime }}=1$. By [L2], $⟨g^p⟩$ has order $p^{n-1}$, including order one at $n=1$, and [L3] is consistent with this cyclic subgroup description.

Formula [L1] gives $\mathrm{\Phi }(P)=⟨g^p⟩$. The quotient has order $p$, so its nonidentity coset $g\mathrm{\Phi }(P)$ is a one-vector basis; hence [F2] gives $d(P)=1$.

Componentwise addition makes $(\mathbb{Z}\mathrm{/}p{)}^n$ a finite abelian group of order $p^n$ by [L2], and every nonzero element has order $p$. It is therefore elementary abelian and [L1] gives $\mathrm{\Phi }((\mathbb{Z}\mathrm{/}p{)}^n)=1$. At $n=0$, the empty product is the trivial group and the same conclusion holds.

The standard coordinate vectors have unique coordinates, so they form a basis by [F1]. There are $n$ of them, hence [F2] gives $d((\mathbb{Z}\mathrm{/}p{)}^n)=n$, including the empty basis at $n=0$.

In $D$, $(r^i{)}^2=r^{2i}$ and $(r^{i}s{)}^2=1$, so $D^2=⟨r^2⟩$. In $Q_8$, the squares are $1$ and $-1$, with every noncentral element squaring to $-1$, so $Q_8^2=\{1,-1\}$. The commutators of [F1] give the same two subgroups: $[r,s]=r(s{r}^{-1}{s}^{-1})=r\cdot r=r^2$ and every commutator of $D$ is a power of $r^2$, so $D^{\mathrm{\prime }}=⟨r^2⟩$; and $[i,j]=ij{i}^{-1}{j}^{-1}=k(-i)(-j)=kij=k^2=-1$, so $Q_8^{\mathrm{\prime }}=\{1,-1\}$.

Apply [L1] to step 1.1. Each quotient has order four and exponent two, with the classes of $r,s$ and of $i,j$ respectively as two-vector bases. Thus both quotients are $(\mathbb{Z}\mathrm{/}2{)}^2$, and [F2] gives generator rank two.

Multiplying $M(a,b,c)M(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})$ entrywise gives the matrix with $(1,2)$ entry $a+a^{\mathrm{\prime }}$, $(2,3)$ entry $b+b^{\mathrm{\prime }}$ and $(1,3)$ entry $c+c^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }}$, so the parametrisation carries matrix multiplication to the displayed coordinate operation and is a bijection. Direct substitution shows associativity; the identity is $(0,0,0)$ and $(a,b,c{)}^{-1}=(-a,-b,-c+ab)$. By [L2] the set has order $p^3$, so this is a finite $p$-group.

A calculation with the inverse in step 1.1 gives $[(a,b,c),(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})]=(0,0,a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b)$. Every commutator lies on the central $c$-axis, and $(0,0,t)=[(t,0,0),(0,1,0)]$, so $H_p^{\mathrm{\prime }}=\{(0,0,c)\}$.

Induction gives $(a,b,c{)}^m=(ma,mb,mc+\left(\genfrac{}{}{0px}{}{m}{2}\right)ab)$. For odd $p$, every $p$th power is the identity; for $p=2$, every square is $(0,0,ab)$ and hence lies on the central axis. Since the derived subgroup already equals that axis, [L1] gives $\mathrm{\Phi }(H_p)=\{(0,0,c)\}$ in both cases.

Modulo the central axis, the first two coordinate classes give unique coordinates and form a two-vector basis by [F2]. Therefore [F1] gives $d(H_p)=2$.

The three displayed subgroups have order four, are distinct, and contain $\mathrm{\Phi }(D)=⟨r^2⟩$ from [L1]. Since $D$ has order eight, each is maximal.

Use the quotient basis $(r\mathrm{\Phi }(D),s\mathrm{\Phi }(D))$. The quotient images of the displayed subgroups are the lines spanned by $(1,0)$, $(0,1)$, and $(1,1)$, which are respectively the kernels of $(a,b)\mapsto b$, $(a,b)\mapsto a$, and $(a,b)\mapsto a+b$. These are precisely the hyperplanes described by [L2].

The $4$-cycle has order four, so $f$ is an embedding. By [L1], $\mathrm{\Phi }(C_4)=⟨g^2⟩$, and $f(g^2)=(13)(24)\mathrm{\ne }1$.

By [L2] and [L6], $\mathrm{\Phi }(S_5)$ is nilpotent and normal. The list [L3] leaves only $1,A_5,S_5$; [L4] and [L5] exclude the latter two, so $\mathrm{\Phi }(S_5)=1$.

Step 1.1 exhibits a nonidentity element of $f(\mathrm{\Phi }(C_4))$, while step 1.2 makes the target Frattini subgroup trivial. Hence $f(\mathrm{\Phi }(C_4))\nleqq \mathrm{\Phi }(S_5)$.

By [L1], $F(A_5)$ is normal and nilpotent. Simplicity [L2] leaves $F(A_5)=1$ or $F(A_5)=A_5$; the second would make $A_5$ nilpotent and hence solvable by [L4], contradicting [L3]. Thus $F(A_5)=1$.

For $A_5$, one has $F(A_5)=1$ and $C_{A_5}(F(A_5))=A_5$, since every element centralizes the trivial subgroup by [F1]. Therefore $C_{A_5}(F(A_5))\nleqq F(A_5)$.

By [L1], $\mathrm{\Phi }(C_3)=⟨g^3⟩=1$. The unit $[-1{]}_3$ gives inversion by [L3]; it sends $g$ to $g^{-1}=g^2\mathrm{\ne }g$ and its square is the identity, so it is a nonidentity automorphism of order two.

Since the Frattini subgroup is trivial, the induced quotient action is the same nontrivial inversion. This is consistent with [L2], which forbids this order-two subgroup from acting trivially because $2$ is coprime to $3$.

Every nonzero vector of $E$ spans one of the three order-two subgroups $⟨(1,0)⟩$, $⟨(0,1)⟩$, and $⟨(1,1)⟩$. Each has index two and is maximal.

Their union is all of $E$, while their intersection is $\{(0,0)\}$, which is $\mathrm{\Phi }(E)$ by [L1] and [F1]. Since $E$ is nontrivial, the union is not the Frattini subgroup.

The embedding in [L1] is a homomorphism for which the claimed inclusion fails, so it refutes the universal statement.

Surjectivity gives the valid replacement. If $f:G\to H$ is onto and $M$ is maximal in $H$, then $f^{-1}(M)$ is maximal in $G$: a subgroup $K$ properly containing it has $f(K)>M$, hence $f(K)=H$; for every $g\in G$, choose $k\in K$ with $f(k)=f(g)$, and then $g{k}^{-1}\in \ker f\le f^{-1}(M)\le K$, so $K=G$. Therefore every $x\in \mathrm{\Phi }(G)$ lies in every $f^{-1}(M)$, and $f(x)$ lies in every maximal $M$, hence in $\mathrm{\Phi }(H)$.

Step 1.1 shows that nonsurjective homomorphisms need not satisfy the inclusion, while step 1.2 identifies the missing sufficient hypothesis.

The singleton $\{[1]\}$ generates $\mathbb{Z}\mathrm{/}6$, and its only proper subset is empty, which generates only $[0]$. Thus it is minimally generating by [F1].

The set $\{[2],[3]\}$ generates because $[2]+[2]+[3]=[1]$. The singleton $\{[2]\}$ generates $\{[0],[2],[4]\}$ and $\{[3]\}$ generates $\{[0],[3]\}$, so neither proper singleton generates; hence the two-element set is also minimal by [F1].

Steps 1.1 and 1.2 exhibit minimal generating sets of sizes one and two in the same finite group, refuting the claim.

For $A_5$, [L1] gives $F(A_5)=1$ and $C_{A_5}(F(A_5))=A_5$.

Thus the claimed inclusion fails for $A_5$. The valid theorem [L2] has finite solvability as a hypothesis, and $A_5$ lies outside that hypothesis.

The strict failure in step 1.1 refutes the hypothesis-free claim.