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In previous readings, I have frequently seen the concept of the subgroup generated by set $X$ (otherwise denoted as $gp(X)$ ) explained as follows:

"$gp(X)$ is the subgroup that is generated from all possible finite compositions of the elements of $X$ and their inverses"

However, I came across a definition today that I have not previously seen. It reads as follows:

"let $gp(X)$ be defined as the intersection of all subgroups of $G$ containing $X$"

Upon first glance, this seems fairly obvious...if a subgroup $H$ contains the elements that comprise the set $X$, then by definition of "subgroup", $H$ clearly also contains the inverse elements of the elements belonging to set $X$. Additionally, because $H$ is a subgroup, it clearly also contains the identity element.

I see that a nice trick to "pick out these elements" is by imposing an intersection with another subgroup $J$ that also contains the set $X$. In this way, subgroup $J$ and subgroup $H$ clearly both contain all elements of set $X$, the inverses of the elements of set $X$, and the identity.

It seems like only two subgroups are necessary to pick out these elements...which inspires two questions:

  1. Why define this as "intersection of all subgroups"?

  2. What happens if there is only one subgroup that contains $X$?

It seems to me that the first defintion avoids these issues.

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    $\begingroup$ 1.) Because this is a universally applicable way to say it. 2.) Then that one subgroup is the subgroup generated by X, which fits the definition without issue. $\endgroup$
    – Jonathan Hebert
    Commented Aug 5, 2019 at 2:14
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    $\begingroup$ They’re equivalent because you prove that they’re equivalent. You do this by showing that every element of the one is in the other, and vice versa. It’s really not hard, you should try it. If you get stuck, ask another question here. $\endgroup$
    – Lubin
    Commented Aug 5, 2019 at 2:29
  • $\begingroup$ See also en.wikipedia.org/wiki/Closure_operator $\endgroup$
    – lhf
    Commented Aug 5, 2019 at 10:38
  • $\begingroup$ @Lubin I guess the issue that I am having in thinking about how to construct a proof for this is that I do not see why I am guaranteed that the intersection of ALL subgroups ensures that there are only elements purely of the form described by the first definition. Obviously, $gp(X)$ is itself a subgroup...and it contains $X$, so as long as this subgroup is present in the intersection, you will certainly only define a group with $gp(X)$ behaving as the "minimum" subgroup that contains $X$. However, this logic seems very circular. $\endgroup$
    – S.C.
    Commented Aug 5, 2019 at 11:54
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    $\begingroup$ Don’t look at it that way, you’re paralyzing yourself. You have two groups: $G_1$ is the intersection of all subgroups of the big overgroup that happen to contain $X$, and $G_2$ is the set of all finite expressions in elements of $X$. Show that every element of $G_2$ is in $G_1$, and then show that $G_2$ is one of the unspecified groups being intersected. It follows that the intersection of all is contained in $G_2$. Then you’re done. If you really can’t fill in the details, I’ll do it for you, but I won’t be happy doing your work. $\endgroup$
    – Lubin
    Commented Aug 5, 2019 at 15:43

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Two arbitrary containing subgroups won't always do. Think about the subgroup of the integers generated by $30$. It's not the intersection of the subgroup generated by $5$ and the one generated by $3$.

If you are allowed to pick the subgroups then just one will do: use the subgroup defined by the other definition.

The alternative definition is equivalent only if you take the intersection of all the subgroups containing the generators.

This kind of equivalent definition pair comes up often. One of the pair describes the object from the inside, the other from the outside.

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    $\begingroup$ Not only that, but in applications, sometimes you want the one characterization, sometimes the other. It’s really to useful to have both characterizations in your toolbox. $\endgroup$
    – Lubin
    Commented Aug 5, 2019 at 2:32
  • $\begingroup$ The implicit composition in your example is addition, correct? $\endgroup$
    – S.C.
    Commented Aug 5, 2019 at 2:51
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    $\begingroup$ @S.Cramer Yes. The integers under multiplication do not form a group, so no one would speak of such subgroups. $\endgroup$
    – Jonathan Hebert
    Commented Aug 5, 2019 at 3:23
  • $\begingroup$ Proving that the 2 definitions are equivalent seems a little circular to me (but maybe it SHOULD look circular if the two definitions are equivalent...?). If I look at ALL subgroups that contain $X$, then clearly $gp(X)$ is one such subgroup. At that point, though, I don't understand why I would even invoke the second definition of the intersection of "ALL" subgroups that contain X. If $gp(X)$ is a subgroup that contains $X$, which it is, then of course the intersection of all subgroups that contain $X$ will yield $gp(X)$. This second definition seems...pointless. $\endgroup$
    – S.C.
    Commented Aug 5, 2019 at 12:12
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    $\begingroup$ @S.Cramer Yes, it may seem circular at first. But you can't argue that $gp(X)$ is the only subgroup you need to consdier in the second definiton because "$gp(X)$" is what you are defining with at the moment. That would be circular reasoning. $\endgroup$
    – Ethan Bolker
    Commented Aug 5, 2019 at 12:49

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