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If $H/N$ is characteristic in $G/N$ and $N$ is characteristic in $G$, then $H$ is characteristic in $G$, a proof could be found here or here. The notation, i.e. speaking about subgroups $H/N$ implies $N \le H$, for general subgroups we must write $HN/N$ to get a subgroup in $G/N$.

So now my question. Do you know an example of a subgroup $N \nleq H$ such that $HN/N$ is characteristic in $G/N$, $N$ is characteristic in $G$, but $H$ is not characteristic in $G$?

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StefanH
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    Yes I know an example! It's easy to find examples in which $H$ is not normal in $G$, but perhaps you would prefer $H$ to be normal. We can still do it! Let $G=D_8$ be dihedral of order $8$. I'll leave you to choose $H$ and $N$. – Derek Holt Mar 29 '15 at 16:56
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    Let $D_8 = \langle x, h \rangle$ with $x^2 = 1, h^4 = 1$. Then $N := \langle h \rangle$ is characteristic in $D_8$ and $D_8 / N \cong \mathbb Z/ 2\mathbb Z$. Also $H = { e, x, h^2, h^2x }$ is normal, but not characteristic, further $HN/N = G/N$, and therefore it is characteristic as it is the whole group. – StefanH Mar 29 '15 at 17:42
  • Yes that's right! – Derek Holt Mar 29 '15 at 17:55

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