Notes for BBS/Martins-150428-112639.jpg
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Some foundational exercises about crossed-modules here!
Note. By $\partial(g\triangleright e)=g\partial(e)g^{-1}$, the image of $\partial$ is a normal subgroup of $G$.
Solution 1. In fact, by the Peiffer relation $\partial(e)\triangleright f=e^{-1}fe$, any $e\in\ker\partial$ is central.
