Group Theory – Does Equality Hold for Subgroups of G?

Question

So it is a well knew result that any intersection of subgroup is a subgroup and even it is a well knew result that the union of subgroup is not generally a subgroup. However, if $\mathcal S(G)$ is the collection of subgroups of any group $G$ then for any subset $X$ of $G$ then the collection
$$
\mathcal G(X):=\{Y\in\mathcal S(G):X\subseteq Y\}
$$
is not empty so that its intersection $\langle X\rangle$ is a subgroup containing $X$ which is moreover the less subgroup containing it: thus if $\mathcal H$ is a collection of subgroups then we can say that
$$
\langle\mathcal H\rangle:=\Big\langle\bigcup\mathcal H\Big\rangle
$$
is the subgroup generated by $\mathcal H$.

Now I know that the equality
$$
\tag{1}\label{1}\begin{equation}\langle\mathcal H\rangle=\Big\{x_1\cdots x_n:x_i\in\bigcup\mathcal H\text{ for all }i=1,\dots, n\text{ with }n\in\omega\Big\}\end{equation}
$$
holds (right?) so that by this I asked if even the equality
$$
\tag{2}\label{2}\begin{equation}\langle G_1\cup G_2\cup G_3\rangle=\big\langle\langle G_1\cup G_2\rangle \cup G_3\big\rangle\end{equation}
$$
holds for any $G_1,G_2,G_3\in\mathcal S(G)$.

So first of all I observed that
$$
G_1\cup G_2\cup G_3\subseteq\langle G_1\cup G_2\rangle\cup G_3\subseteq\big\langle\langle G_1\cup G_2\rangle\cup G_3\big\rangle
$$
holds so that I conclude that
$$
\langle G_1\cup G_2\cup G_3\rangle\subseteq\big\langle\langle G_1\cup G_2\rangle\cup G_3\big\rangle
$$
However, by \eqref{1} any element of $\big\langle\langle G_1\cup G_2\rangle\cup G_3\big\rangle$ is a finite product of elements of $\langle G_1\cup G_2\rangle$ and $G_3$ so that if by \eqref{1} any element of $\langle G_1\cup G_2\rangle$ is a product of elements of $G_1$ and $G_2$ then by associativity and by \eqref{1} we conclude that
$$
\big\langle\langle G_1\cup G_2\rangle\cup G_3\big\rangle\subseteq\langle G_1\cup G_2\cup G_3\rangle
$$
Finally we conclude that the equality \eqref{2} holds.

So I ask if actually \eqref{2} holds because I do not know if really \eqref{1} holds since my textbook do not write it explicitly although, at least it seems, it proves it. Moreover if \eqref{2} holds then is it possibile to prove it using associativity of intersection? For sake of completeness here a reference from my textbook – in Italian, sorry. Could someone help me, please?

Answer 1

Yes, both hold.

For any family $\{H_i\}_{i\in I}$ of subsets (not necessarily subgroups) of $G$, $\langle H_i\rangle$ is the subgroup $K$ that satisfies the following two conditions:

1. $H_i\subseteq K$ for each $i\in I$; and
2. if $M$ is any subgroup of $G$ such that $H_i\subseteq M$ for every $i\in I$, then $K\subseteq M$.

This is the "top-down description" of the subgroup generated. You use these properties in the second displayed equation after $(2)$.

We want to show that $$\bigl\langle G_1\cup G_2\cup G_3\bigr\rangle = \Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle.$$ (We don't even need to assume these are subgroups).

$\subseteq)$ Note that $$\begin{align*} G_1\cup G_2\subseteq \langle G_1\cup G_2\rangle &\subseteq \Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle\\ G_3\subseteq \langle G_1\cup G_2\rangle\cup G_3 &\subseteq \Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle \end{align*}$$ Therefore, $$(G_1\cup G_2)\cup G_3 = G_1\cup G_2\cup G_3 \subseteq \Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle \cup G_3\Bigr\rangle,$$ and since $\langle G_1\cup G_2\cup G_3\rangle$ is the smallest subgroup that contains $G_1\cup G_2\cup G_3$, we have $$\langle G_1\cup G_2\cup G_3\rangle \subseteq \Bigl\langle\bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle.$$

$\supseteq)$ Since $G_1\cup G_2\subseteq G_1\cup G_2\cup G_3\subseteq \langle G_1\cup G_2\cup G_3\rangle$, we have $\langle G_1\cup G_2\rangle \subseteq \langle G_1\cup G_2\cup G_3\rangle$.

We also have $G_3\subseteq G_1\cup G_2\cup G_3\subseteq \langle G_1\cup G_2\cup G_3\rangle$.

Since both $\langle G_1\cup G_2\rangle$ and $G_3$ are contained in $\langle G_1\cup G_2\cup G_3\rangle$, their union is contained there as well, so $$\langle G_1\cup G_2\rangle\cup G_3\subseteq \bigl\langle G_1\cup G_2\cup G_3\bigr\rangle,$$ and therefore $$\Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle \cup G_3\Bigr\rangle \subseteq \bigl\langle G_1\cup G_2\cup G_3\bigr\rangle,$$ as desired. $\Box$

More generally, for any family $\{B_i\}_{i\in I}$ of subsets of $G$, and any partition $\{I_j\}_{j\in J}$ of $I$ (so $\cup_{j\in J}I_j=I$), we have $$\left\langle \bigcup_{i\in I}B_i\right\rangle = \left\langle \bigcup_{j\in J}\left\langle \cup_{i\in I_j}B_k\right\rangle\right\rangle.$$

As to whether your description of $(1)$ is accurate, note that in general if $S$ is a subset of $G$, $\langle S\rangle$ is the collection of all finite products of elements of $S$ and their inverses. But if $S$ is closed under inverses, that is, if $S$ satisfies the condition $$\text{if }s\in S,\text{ then }s^{-1}\in S,$$ then this is exactly the same thing as the collection of all finite products of elements of $S$. (Note the empty product is a finite product and gives the identity element). Since your $\cup\mathcal{H}$ is a union of subgroups, it is certainly closed under inverses; and you can ignore the empty product since the set already contains $e$; so your description of $\langle \cup\mathcal{H}\rangle$ is accurate if the family of subgroups is nonempty.