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By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, it$V^N$ is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent). Therefore G is unipotent.

By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, it is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent).

By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, $V^N$ is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent). Therefore G is unipotent.

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anon
anon
  • 151
  • 2

By definition, G is unipotent if and only if every nonzero representation has nonzero fixed vectors. Consider a representation V of G. As N is unipotent, $V^N$ is nonempty. Because N is normal, it is stable under G, hence under G/N, and hence has nonzero fixed vectors (because G/N is unipotent).