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A hereditary property is a property of an object that is inherited by all of its subobjects where the meaning of sub-object depends on the context.

Particularly the hereditary property is of great interest because we use it most of the time in our arguments, in theorems. It is always nice to observe these subtle properties which make our life simpler. If I know that some property is hereditary then it is of great advantage for students while solving problems.

This hereditary property is used in almost every branch of mathematics but to make the context narrow we focus only on abstract algebra. I am looking to identify whether a property is hereditary or not in abstract algebra especially in groups, rings, and fields.

Here are some examples that are familiar to me :

  1. There exists a group with infinite order such that at one least non-trivial subgroup has finite order(for example consider non-zero real numbers under multiplication). So this property is not hereditary. Whereas every subgroup of the finite group has the finite order so this property is hereditary.
  2. Abelianness of a group is hereditary whereas non-Abelianness is not.
  3. Since every subgroup of a cyclic group is cyclic this property is hereditary.
  4. If $H$ is a normal subgroup of $G$ then every subgroup of the factor group $G/H$ is in the form $N/H$ where $N$ is a subgroup of $G$.
  5. Every subring of an integral domain is again an integral domain. So this a hereditary property.
  6. If $R$ is a principal ideal domain then there may be a subdomain of $R$ that need not be a principal ideal domain since $\mathbb{C}$ is a PID while $\mathbb{Z}[\sqrt{5}]$ is not a PID. So this is not a hereditary property. (After a comment from @ GreginGre)
  7. The property of being the norm Euclidean domain is not hereditary because $\mathbb{Z}[\sqrt 2]$ is the norm Euclidean whereas its subdomain $\mathbb{Z}[\sqrt {98}]$ is not.
  8. Let $F$ be a finite field of order $p^n$ for some prime $p$ and a natural number $n$. Every subfield of $F$ has order $p^m$ for some $m < n$. So this property is hereditary.

It will be nice if you add about whether a property is hereditary or not along with a proper reason.

Infinity_hunter
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    Are you saying $\mathbb Z[\sqrt{14}]$ is a subdomain of $\mathbb Z[\sqrt2]$?! – J. W. Tanner Jun 22 '21 at 08:06
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    @ J. W. Tanner Sorry, it should be $98$ instead of $14$, it was a typo. – Infinity_hunter Jun 22 '21 at 08:15
  • Being a PID is NOT hereditary. $\mathbb{C}$ is a PID while $\mathbb{Z}[\sqrt{5}]$ is not. – GreginGre Jun 22 '21 at 11:09
  • @rschwieb Ok I agree that it is not very clear what it implies. Let's not have any confusion, I will not remove that. – Infinity_hunter Jun 22 '21 at 12:42
  • @Infinity_hunter I figured you did not mean that :) looks good now – rschwieb Jun 22 '21 at 13:05
  • If anything, please provide motivation for why you consider a hereditary property worth collecting. For example, have you seen the use of a "hereditary property" in a particular proof? Do you think it holds significance as an algebraic exercise to collect examples of hereditary properties? It should be on you to motivate this : providing these details ,and asking people to elaborate on their reasons for including a certain theorem, will make for a more informed and comprehensive post. This is a big-list question, and I think a big-list question should have these details from the OP. – Sarvesh Ravichandran Iyer Jun 22 '21 at 13:16
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    In case you want concrete motivation, you could point to "Markov properties" and the Adian-Rabin Theorem, which essentially says that hereditary properties in groups not algorithmically detectable. For example, given a group via a presentation it is undecidable if the group is trivial, cyclic, abelian, etc., because these are all hereditary properties. See wikipedia, or these two answers (of mine): 1, 2. – user1729 Jun 22 '21 at 13:49
  • @Teresa Lisbon I have edited my post and added why this property is of interest. Is that fine? – Infinity_hunter Jun 22 '21 at 13:53
  • @ user1729 It will be great if you add this as an answer – Infinity_hunter Jun 22 '21 at 14:00
  • @Infinity_hunter Certainly. I mean , an explicit example of a theorem assisted by the property would have been the best : adding user1729's links (if you've read them you can add further visited links/results that you saw as well) will make this question reign supreme in any case. I had a downvote, I've removed it : thanks for your efforts in this positive direction. – Sarvesh Ravichandran Iyer Jun 22 '21 at 14:56
  • Any property defined by an identity would necessarily be hereditary in this sense, as would any property defined by a quasiidentity. – Arturo Magidin Jun 22 '21 at 17:31
  • @Infinity_hunter My comment isn't a good fit as an answer, but instead is a motivation. You could add this to your question, although it doesn't really address the "too broad" close reason (so maybe if you asked about hereditary properties of groups instead then you could include my comment - I'm happy to help with this). – user1729 Jun 23 '21 at 11:49

2 Answers2

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Here are some interesting ones:

  • If $N \le H \le G$ are groups such that $N$ is normal in $H$ and $H$ is normal in $G$, then it need not be true that $N$ is normal in $G$.

  • If $I \subseteq J \subseteq R$ is an inclusion of subrings (not necessarily unital) where $I$ is an ideal in $J$ and $J$ is an ideal in $A$, then $I$ is not necessarily an ideal in $R$. However, if $R$ has the structure of a $C^*$-algebra this is true.

  • A subgroup of a free group is free. This is non-trivial.

J. De Ro
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Associativity is hereditary in any group, distributivity is hereditary in any ring/vector field (even if the chosen subset is not a substructure)

Samuel M. A. Luque
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