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A group G satisfies the normalizer condition (or G is an N-group) if every proper subgroup of G is a proper subgroup of its normalizer (i.e. no subgroup is self-normalizing). At first I took for granted that this implies that every subgroup is also an N-group, but then I realized that it's not trivial: if H is a subgroup of G and K is a subgroup of H, then K is strictly contained in its normalizer in G, but that doesn't mean it's also strictly contained in its normalizer in H. I tried to find a counter-example but I couldn't...

Does any of you have an example of an N-group which has a subgroup which isn't an N-group? Also, can we find a sufficient condition on H (subgroup of an N-group G) so that H is an N-group? (I was thinking about H normal in G, but I don't seem to be going anywhere...).

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  • $\begingroup$ So I assume you are familiar with how the proof goes for finite groups. Not sure if that really gives any indication what the answer should be in general though. $\endgroup$ – Tobias Kildetoft Jan 7 '17 at 18:02
  • $\begingroup$ @TobiasKildetoft No, I'm not... $\endgroup$ – frafour Jan 7 '17 at 18:07
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    $\begingroup$ For finite groups this property is equivalent to being nilpotent. $\endgroup$ – Tobias Kildetoft Jan 7 '17 at 18:30
  • $\begingroup$ @TobiasKildetoft That's right, thanks! $\endgroup$ – frafour Jan 7 '17 at 18:37

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