McLain's characteristically simple group

Question

Given a field $\mathbb{F}$, one can construct the free $\mathbb{F}$-module over the set of rational numbers. Assume that $(v_x)_{x \in \mathbb{Q}}$ is a basis of this linear space (indexed by the rational numbers). For each pair of rational numbers $x <y$ consider $e_{xy}$ the linear transformation that maps $v_x$ to $v_y$, and maps the other elements of the basis to $0$. Now we define the McLain group $M=\operatorname{M}(\mathbb{Q},\mathbb{F})$ to be the group generated by all the linear transformations of the form $1+ae_{xy}$ , $x<y$ and $a \in \mathbb{F}$ (note that all of these transformations are invertible).
A description of such a group is in Robinson's "A course in the Theory of Groups", 12.1.9 or in the paper of Philip Hall or in the original work of McLain or in Robinson's ``Finiteness Conditions and Generalized Soluble Groups'', Vol. 2, p.14.

I was trying to prove that the McLain group is directly indecomposable, but I cannot succeed.

Let's say that $M=H\times K$ ($H$ and $K$ both non-trivial and normal subgroups of $M$) with $[H,\, K]=1$, I know that in each normal subgroup one can find a generator $1+ae_{xy}$ but I cannot see how to relate these two normal subgroups.

Any idea will be appreciated.

Note that the fact that $M$ is directly indecomposable is just a guess of mine, so I can be wrong.

Answer 1

Note that

$$ [1+\gamma e_{xy}, 1+\delta e_{yz}] = (1-\gamma e_{xy})(1-\delta e_{yz})(1+\gamma e_{xy})(1+\delta e_{yz}) = 1 + \gamma\delta e_{xz}. $$

It is stated in the question that if $L$ is a normal subgroup of $G$ then $L$ contains some $1+\alpha e_{xy}$. By taking commutators with $1+\beta e_{wx}$ and $1+\beta e_{yz}$ for $w < x$ and $z > y$, it follows that $L$ contains all $1 + e_{wz}$ with $w \le x < y \le z$. Hence any two normal subgroups of $G$ intersect.

(The first version of this answer was wrong because I overlooked the condition $x < y$ in the definition of the McLain group.)