This is part of Exercise 5.2.5 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.
Here are some previous questions of mine about Wreath products:
Showing $Z(H\wr K)=1$, for abelian $H\neq 1$ and arbitrary, infinite $K$.
Showing that the restricted wreath product $\Bbb Z\wr\Bbb Z$ is finitely generated.
The Details:
(This can be skipped.)
On page 32 and 33, ibid.,
Let $H$ and $K$ be permutation groups on sets $X$ and $Y$ respectively. [. . .]
If $\gamma\in H$, $y\in Y$, and $\kappa\in K$, define
$$\gamma(y):\begin{cases} (x,y) \mapsto (x\gamma, y) & \\ (x,y_1) \mapsto (x,y_1) & y_1\neq y,\end{cases}$$ and
$$\kappa^\ast: (x,y)\mapsto (x,y\kappa).$$
[. . .] The functions $\gamma\mapsto \gamma(y)$, with $y$ fixed, and $\kappa\mapsto \kappa^\ast$ are monomorphisms from $H$ and $K$ to ${\rm Sym}(X\times Y)$: let their images be $H(y)$ and $K^\ast$, respectively. Then the wreath product of $H$ and $K$ is the permutation group on $X\times Y$ [. . .]
$$H\wr K=\langle H(y), K^\ast\mid y\in Y\rangle.$$
[. . .]
[T]he base group $B$ of the wreath product [is]
$$B=\underset{y\in Y}{{\rm Dr}}\, H(y).$$
Here $\underset{y\in Y}{{\rm Dr}}\, H(y)$ is defined as in this question.
On page 41 of the book,
If $H$ and $K$ are arbitrary groups, we can think of them as permutation groups on their underlying sets via the right regular representation and form their wreath product $W=H\wr K$: this is called the standard wreath product.
On page 94 of the book,
[Let $p$ be prime and let] $P$ be the group with the generators $a_1,a_2,a_3,\dots$ and relations
$$pa_1=0,\quad pa_{i+1}=pa_i,\quad a_i+a_j=a_j+a_i,$$
[then] $P$ is the [. . .] quasicyclic $p$-group.
On page 130 of the book,
[B]y a maximal subgroup we mean a proper subgroup which is not contained in any larger proper subgroup.
On page 135 of the book,
The Frattini subgroup of an arbitrary group $G$ is defined to be the intersection of all maximal subgroups, with the stipulation that it shall equal $G$ if $G$ should prove to have no maximal subgroups. This subgroup [. . .] is written
$${\rm Frat}\, G.$$
. . . and . . .
[A]n element $g$ is called a nongenerator of $G$ if $G=\langle g, X\rangle$ always implies that $G=\langle X\rangle$ when $X$ is a subset of $G$.
The Question:
Let $H$ and $K$ be quasicyclic groups and write $G=H\wr K$ for the standard wreath product. Prove that $G={\rm Frat}\, G$.
Thoughts:
Clearly $G$ is infinite.
By definition of ${\rm Frat}\, G$, we need to show $H\wr K$ has no maximal subgroups.
Suppose the prime that defines $H$ is $p$ and that of $K$ is $q$.
The following is on page 135.
Theorem 5.2.12: (Frattini). In any group $G$ the Frattini subgroup equals the set of nongenerators of $G$.
Therefore, if every $g$ of $H\wr K$ is a nongenerator, then we are done.
Suppose $g\in G$ and $G=\langle g, X\rangle$ for some $X\subseteq G$. WLOG, $g\notin X$. I need to show $H\wr K=\langle X\rangle$. We have the presentation in the details above, which might help, but I'm not sure.
I'm not particularly comfortable with quasicyclic groups. However, the following defines a multiplicative group isomorphic to the quasicyclic $p$-group:
$$\Bbb Z(p^\infty)=\{z\in\Bbb C\mid z^{(p^n)}=1\text{ for some }n\in\Bbb N\},$$
which can be found here and in Kargapolov et al.'s "Fundamentals of the Theory of Groups". I find it much easier to understand.
Suppose $M\le G$ is maximal. I can see, sort of, that $M$ cannot exist, because there is no limit on $n$ in the "$z^{(p^n)}=1$" part of the equivalent definition above.
I don't think I could answer this myself any time soon.
Please help :)
