On $p$-groups with a unique minimal subgroup

Question

If $G$ is a finite group with a unique minimal subgroup, we know that $|G|=p^n$. I have to prove that if $p\neq2$ then $G$ is cyclic.

This is the contest. What I don't understand is the following part (in the proof):

we know that $G/\Phi(G)\simeq C_p^{d}$ where $\Phi(G)$ is the Frattini subgroup and $C_p^{d}$ are $d\in\mathbb N$ copies of the cyclic group of order $p$; moreover we know that $d$ is the minimum cardinality of a set of generators for $G$. Hence $M\le G$ maximal $\Longrightarrow$ $\#$generators of $M= d-1$ (if it were $<d-1$ $M$ wouldn't be maximal, if it were $=d$ then $M=G$). Then my teacher wrote $M/\Phi(G)\simeq C_p^{d-1}$, but I think it's a mistake, because, for what I wrote just now, it must be $M/\Phi(M)\simeq C_p^{d-1}$.

Am I right? Thank you all.

Answer 1

Here is a reasonably important example to keep in mind when thinking about generating sets of $p$-groups. For any finite $p$-group $G$, define $d(G)$ to be the solution to $|G/\Phi(G)| = p^d$. It is well-known that $d(G)$ is the size of every minimal generating set of $G$ (this is one statement of Burnside's basis theorem).

Let $\tau_{p,n}$ be the permutation of order $p$ on $p^n$ points that is $p^{n-1}$ cycles of length $p$, with the $i$th cycle being $\left(i,~p^{n-1} +i,~2p^{n-1}+i, \ldots, ~(p-1)p^{n-1}+i\right)$.

Let $G_{p,n} = \langle \tau_{p,1}, \tau_{p,2}, \ldots, \tau_{p,n} \rangle$ be a Sylow $p$-subgroup of the symmetric group on $p^n$ points. It is not too hard to prove that $d(G_{p,n}) =n$ exactly, but it is clear that $d(G_{p,n}) \leq n$.

Now let $H_{p,n}$ be the group generated by the $p^{n-1}$ commuting $p$-cycles $\sigma_i = (i, p+i,2p+i, (p-1)p+i)$. Each $\sigma_i$ is a $G_{p,n}$ conjugate of $\tau_{p,1}$, so $H_{p,n} \leq G_{p,n}$.

However, $\Phi(H_{p,n}) = 1$, so $d(H_{p,n})=p^{n-1}$ exactly.

For ($p>2$ and $n>1$) or ($p=2$ and $n>2$), $p^{n-1} > n$, so we have many examples of $H=H_{p,n}$ and $G=G_{p,n}$ with $H \leq G$ and $d(H) > d(G)$.

Specific case of 8

Here is $p=2$ and $n=3$: $\tau_{2,1} = (1,2)$, $\tau_{2,2}=(1,3)(2,4)$, $\tau_{2,3}=(1,5)(2,6)(3,7)(4,8)$. $G=G_{2,3} = \langle (1,2), (1,3)(2,4), (1,5)(2,6)(3,7)(4,8) \rangle$.

It is not too hard to compute the maximal subgroups etc. to compute $\Phi(G)$, but it is immediately clear that $d(G) \leq 3$.

$H=H_{2,3} = \langle (1,2), (3,4), (5,6), (7,8) \rangle$ clearly has $\Phi(H)=1$ and $d(H)=4$ since $|H|=2^4$. Note that $(1,2)^{(1,3)(2,4)} = (3,4)$ so that $H$ really is a subgroup of $G$.

We have $H \leq G$ but $4 \not\leq 3$.

Specific case of 9

Here is $p=3$, $n=2$: $\tau_{3,1} = (1,2,3)$, $\tau_{3,2}=(1,4,7)(2,5,8)(3,6,9)$, $G=G_{3,2} = \langle (1,2,3), (1,4,7)(2,5,8)(3,6,9) \rangle$. Since $G$ is not cyclic but is generated by two elements, it is clear that $d(G) = 2$ exactly (and immediately clear that $d(G) \leq 2$).

$H=H_{3,2} = \langle (1,2,3), (4,5,6), (7,8,9) \rangle$ clearly has $\Phi(H)=1$ so $d(H) = 3$ since $|H|=3^3$.

Again we have $H \leq G$ but $3 \not\leq 2$.