Can non-abelian subgroups be found in abelian groups?

Question

I am aware that abelian-subgroups can be found in non-abelian groups. All one must do is consider $gp(\{x\})$.

However, this had me wondering if the other way around was true. I couldn't find anything on the forum that asked this… so here's hoping someone could shed some insight.

If the answer is no, it is sort of interesting because, in a hand-wavey way, I interpret this as meaning that a non-abelian subgroup effectively "taints" the overarching group, stripping it of the ability to be abelian.

This makes me wonder why abelian subgroups do not taint non-abelian groups from being non-abelian. Why does the “abelian property” behave so sensitively while the “non-abelian property” does not?

How could that possibly work. If $a,b \in G \subset H$ and $H$ is abelian then $ab =ba$ so $G$ is abelian. Nothing to do with tainting. — fleablood, Aug 05 '19 at 23:23
In the jargon, we'd say that abelian-ness is a hereditary property of groups. — Kaj Hansen, Aug 05 '19 at 23:32
"I couldn't find anything on the forum that asked this" - I could - see here. — Dietrich Burde, Aug 06 '19 at 08:40

score 10 · Accepted Answer · edited Aug 05 '19 at 23:09

The answer is negative: If $G$ is an Abelian group and $H$ is a subgroup of $G$, then $H$ is Abelian too. In fact, since$$(\forall g_1,g_2\in G):g_1g_2=g_2g_1,$$then, in particular,$$(\forall g_1,g_2\in H):g_1g_2=g_2g_1.$$

And note that being Abelian means that something always occurs. And being non-Abelian means that that thing sometimes doesn't occur. For such a property, it is clear that if a group has it, every subgroup also has it, but not the other way around.

score 1 · Answer 2 · answered Aug 05 '19 at 23:24

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How could that possibly work?

If $a,b \in G \subset H$ and $H$ is abelian then $a*b =b*a$ so $G$ is abelian.

The "tainting" issue should be obvious. If $X$ is a condition that SomethingCanBeSaidAboutEveryElement (or pair of elements) it'd be inheritable by subsets as the elements of subset are subsets of the original. But if $X$ is an "anti" condition that it isn't true for all elements it takes just one counter example to achieve this. It need not be inherited by a subset that avoids the counter issue. But supersets will inherrit it.

It's like being sick.. An organ is healthy if none of its components is sick. So if a body is healthy it has no sick components. So none of the components are sick. But if you have one sick cell you are not healthy but if you take a component that excludes the cell that may be healthy. But every component using that cell will be sick.

Or being .... blue. If we say a set is Blue if all its elements are blue, but a set is NotBlue if at least one element is not blue then.....

Or... maybe I dont understand your question.

answered Aug 05 '19 at 23:24

fleablood

124,253

I guess I thought of the physical concept of "emergent properties" and wondered if there was some sort of mathematical equivalent. In this particular instance, I thought, "hmmmm, well, at the small level, abelian properties are not present...but at a large scale level, maybe abelian properties ARE present". – S.C. Aug 05 '19 at 23:31
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But "abelianness" is a global property. Either every it applies to every pair of elements. Or the group isn't abelian. This... isn't philosophy. It's just common sense. A community is smoke free if no-one smokes. It's tainted if a single person smokes. If a country is smoke free it passes down to it's districts, counties, cities, neighbor hoods, streets and houses. But if there is a smoker nothing including it is smoke free. But it's nextdoor neighbor maybe. No philosophy. – fleablood Aug 06 '19 at 00:54
Well, I don't know if emergent properties are "philosophy" (they are literally used to describe chemical systems, biology, etc), but I understand your point. Cheers~ – S.C. Aug 06 '19 at 01:01

Can non-abelian subgroups be found in abelian groups?

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