= Solution
No. Let $Q$ be the <quiver> $1\to2$ over a field $k$, and take the <representation of a quiver>
$$
k\xrightarrow{1_k}k.
$$
An <endomorphism> is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its <endomorphism ring> is $k$. Every nonzero endomorphism is therefore an <isomorphism>, making this representation a <brick module>. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an <irreducible module>.
Equivalently, this is a nonsimple module over the <path algebra> $kQ$ whose endomorphism ring is a division ring.
Solved by gpt-5.6-sol high.
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