Notes for BBS/Martins-150428-112639.jpg: Difference between revisions
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'''Note.''' By $\partial(g\triangleright e)=g\partial(e)g^{-1}$, the image of $\partial$ is a normal subgroup of $G$. |
'''Note.''' By $\partial(g\triangleright e)=g\partial(e)g^{-1}$, the image of $\partial$ is a normal subgroup of $G$. |
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'''Solution 1.''' In fact, by the Peiffer relation $\partial(e)f=e^{-1}fe$, any $e\in\ker\partial$ is central. |
'''Solution 1.''' In fact, by the Peiffer relation $\partial(e)\triangleright f=e^{-1}fe$, any $e\in\ker\partial$ is central. |
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Latest revision as of 17:57, 2 May 2015
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.
