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Problem 32 (Relationship Between S And Its Span)

Let $S$ be a set of permutations of $X$.

  1. Prove that $S\subseteq \text{span}(S)$. In other words, we know that $S$ is always contained in its span; the span might be larger.
  2. Prove that if $\text{span}(S)\subseteq S$, then $S$ is closed.
  3. If $T\subseteq S$, then show that $\text{span}(T)\subseteq\text{span}(S)$.

Solution

Suppose $X$ is a set. Let $S$ be a set of permutations of $X$. We will prove that $S \subseteq \span(S)$. Let $x\in S$. Then $x=x^1\in \span(S)$, which means $S\subseteq \span(S)$.

We now claim if $\span(S)\subseteq S$, then $S$ is closed. To prove this, we let $\span(S)\subseteq S$ and recall from the first part of this proof that $S\subseteq \span(S)$. Thus by definition, we have shown that $S$ is closed.

Finally, we claim that if a set $T\subseteq S$, then $\span(T)\subseteq \span(S)$. Let $m\in \span(T)$ This means $m=t_1^{n_1} \circ t_2^{n_2} \circ \cdots \circ t_k^{n_k}$ for some $k\in \mathbb{N}$ where $t_i\in T$ and $n_i\in \mathbb{Z}$ for each $i$ from 1 to $k$. For each relevant $i$ each $t_i \in S$ because $T\subseteq S$. This means $m\in \span(S)$. Thus $\span(T)\in \span(S)$. $\square$

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