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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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    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. – 2017-01-07
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    @TobiasKildetoft No, I'm not... – 2017-01-07
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    For finite groups this property is equivalent to being nilpotent. – 2017-01-07
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    @TobiasKildetoft That's right, thanks! – 2017-01-07

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