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The Krull–Schmidt theorem says:
If $G$ is a group that satisfies ACC and DCC on normal subgroups, then there is a unique way of writing $G$ as a direct product of finitely many indecomposable subgroups of $G$.
My question is:
If a group $G$ is not simple, doesn't this mean that $G$ is decomposable according to Krull–Schmidt Theorem?
The problem is that you are confusing having a normal proper normal subgroup with being decomposable. A group is decomposable if and only if it has a proper normal subgroup with a normal complement: that is, there have to exist normal subgroups $N$ and $M$, both proper, such that $N\cap M=\{1\}$ and $NM=G$. But in order to be non-simple, you only require the existence of a proper normal subgroup $N$.
Not every normal subgroup has a complement: this is not true even in abelian groups! Not every embedding $f\colon B\to A$ of a subgroup into an abelian group has a retraction (a homomorphism $g\colon A\to B$ such that $g\circ f = \mathrm{id}_B$), which is the condition required for $B$ to be a direct summand/direct factor of $A$.
This is exactly the same phenomenon you see when trying to diagonalize a linear transformation on a vector space of dimension $2$: in order for $T\colon\mathbf{V}\to\mathbf{V}$ to be diagonalizable you need to find two distinct one-dimensional subspaces that are $T$-invariant (proper $T$-invariant subspace play the role of normal subgroups above). For diagonalization to be impossible, one of two things must happen:
There are no 1-dimensional $T$-invariant subspaces (this would be the parallel of the group being simple); this can happen over the real numbers, for example with a rotation by $90^{\circ}$.
But it's possible for $T$ to not be diagonalizable, and yet for there to be a $1$-dimensional $T$-invariant subspace; but not two. An example is the linear transformation $T(x,y) = (x+y,y)$. The subspace $\{(x,0)\mid x\in F\}$ is $T$-invariant, but it's the only $T$-invariant subspace; there is no "second subspace" that you need to diagonalize. (This is the analogue of the cyclic group of order $p^2$, which has a unique proper nontrivial subgroup).
You seem to think that only the first case above can occur (no $T$-invariant subspace/ no normal subgroup). You are ignoring the possibility of the second case (there are $T$-invariant subspaces, just not enough of them/ there are normal subgroups, but none has a normal complement).
$\mathbb Z/p^2\mathbb Z$ is indecomposable but not simple.
The best example of a finite indecomposable group which is not simple is $\mathbb{S}_n$ for $n\geq 3$ (Permutation group). This is easy to prove as well.
