Frattini subgroup of $\operatorname{Hol}(\mathbb Q)$ Let $\mathbb Q$ be rational number under addtion. Is Frattini subgroup of $\operatorname{Hol}(\mathbb Q)$ trivial where $\operatorname{Hol}(\mathbb Q)=\mathbb Q \rtimes \operatorname{Aut}(\mathbb Q) $?
 A: It seems to me that the Frattini subgroup is indeed trivial.
Note that $\operatorname{Aut}(\mathbb Q) = \mathbb{Q}^{\star}$, acting on $\mathbb{Q}$ by multiplication. 
Since $\mathbb{Q}^{\star}$ acts transitively on the nonzero elements of $\mathbb{Q}$, it follows that $\mathbb{Q}$ is minimal normal in $G = \operatorname{Hol}(\mathbb Q)$. This shows that $\mathbb{Q}^{\star}$ is a maximal subgroup of $G$.
Now note that $G$ is isomorphic to the matrix group
$$
H = \left\{ \begin{bmatrix} 1&0\\a&b \end{bmatrix} : a \in \mathbb{Q}, b \in \mathbb{Q}^{\star} \right\},
$$
where $\mathbb{Q}^{\star}$ corresponds to $\left\{ \begin{bmatrix} 1&0\\0&b \end{bmatrix} :  b \in \mathbb{Q}^{\star} \right\}$, and $\mathbb{Q}$ corresponds to $\left\{ \begin{bmatrix} 1&0\\a&1 \end{bmatrix} :  a \in \mathbb{Q} \right\}$.
Now consider the conjugate subgroup of $\mathbb{Q}^{\star}$ by $\begin{bmatrix} 1&0\\a&1 \end{bmatrix}$, for a fixed $a \ne 0$ (you may take $a=1$, for simplicity). Since
$$
\begin{bmatrix} 1&0\\a&1 \end{bmatrix}^{-1}
\begin{bmatrix} 1&0\\0&b \end{bmatrix}
\begin{bmatrix} 1&0\\a&1 \end{bmatrix}
=
\begin{bmatrix} 1&0\\-a&1 \end{bmatrix}
\begin{bmatrix} 1&0\\0&b \end{bmatrix}
\begin{bmatrix} 1&0\\a&1 \end{bmatrix}
=
\begin{bmatrix} 1&0\\a(b-1)&b \end{bmatrix},
$$
this conjugate is
$$
\left\{ \begin{bmatrix} 1&0\\a(b-1)&b \end{bmatrix} : b \in \mathbb{Q}^{\star} \right\}.
$$
This is again a maximal subgroup of $G$, and intersects $\mathbb{Q}^{\star}$ trivially.
Therefore the Frattini subgroup of $G$ is trivial.
