I have spent days now trying to understand the proof of an "easy lemma" in Martin Isaacs' book "Algebra: A graduate course". The lemma is 11.10 on page 150, which reads the following:
Let $N\subseteq M$, where $M$ is a completely reducible abelian $X$-group. Then $N$ and $M/N$ are completely reducible.
I understand the proof except for one small step, which reads as follows:
Now $N=M \dotplus L$ for some $X$-subgroup $L\subseteq N$...
Recall that if $G=A \dotplus B$ is an internal direct sum, then $A$ and $B$ are subgroups of $G$. Internal direct products (and sums) always concern subgroups of a group.
My question:
Since the direct sum $N=M \dotplus L$ in the lemma is an internal direct sum, the summands $M$ and $L$ must be $X$-subgroups of $N$. But it is $N$ which is a subgroup of $M$, not the other way round. So it makes no sense to me that you take the group $M$ and obtain from it a (possibly proper) subgroup $N$ by taking an internal direct sum. How can an internal direct sum of 2 abelian groups be a proper subgroup of one of the summands?
What am I missing? Thank you in advance.