Subgroup lattice

Section 1.10 Subgroup lattice

A deep understanding of a group \(G\) will include an understanding of its subgroup structure. In more detail we want to know (a) what the set \(\mathcal{S}\) of all subgroups of \(G\) is, and (b) for all \(H_1, H_2\in \mathcal{S}\text{,}\) we want to know whether \(H_1\leq H_2\text{,}\) \(H_2\leq H_1\text{,}\) or neither.

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Using slightly fancier language, the pair \((\mathcal{S}, \leq)\) of all subsgroups of \(G\) partially ordered by the set inclusion relation \(\leq\) (which recall is just \(\subseteq\) for subgroups) is what is called a partially ordered set, or lattice. We wish to understand the precise structure of this lattice.

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Proposition 1.10.1. Intersection of subgroups.

Let \(G\) be a group, and let \((H_i)_{i\in I}\) be a family of subgroups \(H_i\leq G\) indexed by the nonempty set \(I\text{.}\) The intersection \(H=\bigcap_{i\in I}H_i\) is a subgroup of \(G\text{.}\) In plain English, the intersection of a family of subgroups is again a subgroup.

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Proof.

Note that the indexing set \(I\) is arbitrary here: it need not be finite, or even countable. This does not complicate our proof in anyway: in the end, this is an exercise in dealinging with the quantifier “for all” that defines the intersection. Let’s show that the three axioms of Definition 1.8.1 are satisfied.

Since \(e\in H_i\) for all \(i\in I\text{,}\) by definition \(e\in \bigcap_{i\in I}H_i\text{,}\) and thus \(e\in H\text{.}\)
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Assume \(g,g'\in H\text{.}\) By definition, this means \(g,g'\in H_i\) for all \(i\in I\text{.}\) Since each \(H_i\) is a subgroup of \(G\text{,}\) we have \(gg'\in H_i\) for all \(i\in I\text{,}\) and thus \(gg'\in H\text{.}\)
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Assume \(g\in H\text{.}\) By definition, this means \(g\in H_i\) for all \(i\in I\text{.}\) Since each \(H_i\) is a subgroup of \(G\text{,}\) we have \(g^{-1}\in H_i\) for all \(i\in I\text{,}\) and thus \(g^{-1}\in H\text{.}\)
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Definition 1.10.2. Subgroup generated by a set.

Let \(A\) be a subset of the group \(G\text{.}\) The subgroup generated by \(A\), denoted \(\angvec{A}\text{,}\) is defined as the intersection of all subgroups of \(G\) containing \(A\text{:}\) i.e.,

\begin{equation} \angvec{A}=\bigcap_{\substack{H\leq G \\ A\subseteq H}} H \text{.}\tag{1.10.1} \end{equation}

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Proposition 1.10.3. Subgroup generated by a set.

Let \(A\) be a subset of \(G\text{.}\)

\(\angvec{A}\) is the smallest subgroup (with respect to the \(\leq\) ordering) of \(G\) containing \(A\text{:}\) i.e., if \(H\leq G\) and \(A\subseteq H\text{,}\) then \(\angvec{A}\subseteq H\text{.}\) Using logical shorthand:

\begin{equation} A\subseteq H\leq G\implies \angvec{A}\leq H\text{.}\tag{1.10.2} \end{equation}

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\(\angvec{A}\) is the set of all finite products of elements of \(A\) and their inverses:

\begin{equation} \angvec{A}=\{a_1^{\epsilon_1}a_2^{\epsilon_2}\cdots a_n^{\epsilon_n}: n\in \Z_{\geq 0}, a_i\in A, \epsilon_i\in \{-1,1\}\}\text{.}\tag{1.10.3} \end{equation}

We declare, by convention, that a product of \(n=0\) elements (the empty product) is equal to \(e\text{.}\)

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