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?

Answer 1

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.

Answer 2

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.