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I'm reading J.S. Milne's notes on Group Theory (highly recommended) and am attempting to understand the following;

Let $X$ be a set of symbols $X = \left\lbrace a, b, c, \dots \right\rbrace$ (possibly infinite) and let $S_X$ be the free monoid on $X$ (binary operation given by concatenation of words). When we identify a symbol $a \in X$ with a word $a \in S_X$, $X$ becomes a subset of $S_X$ and generates it (i.e. no proper submonoid of $S_X$ contains $X$).

I'm not entirely sure what is meant here. When he says "identify $a\in X$ with $a \in S_X$" does he mean "when we treat $a$ as a member of $S_X$"? Also, what is meant by "$X$ becomes a subset of $S_X$"? I understand that $S_X = \langle X\rangle$ because $S_X$ is built from the symbols in $X$ but I'm not sure how to justify that no proper submonoid of $S_X$ contains $X$.

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    $\begingroup$ Different data types: $a\in X$ is a member of a set, $a\in S_X$ is a word of a monoid. $\endgroup$
    – Wuestenfux
    Jun 13, 2017 at 14:47
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    $\begingroup$ Technically, one could regard a symbol $a$ in $X$ and the one-symbol word $a$ as different objects. Milne is saying no, consider them to be the same thing. It's the same thing you do when you're constructing $\mathbf{Z}$ for example. An integer is really an equivalence class (for a certain equivalence relation) of pairs $(a,b)$ where $a$ and $b$ are natural numbers. But once the construction is complete, you say "Let's identify the natural number $n$ with the equivalence class of $(n,0)$." Technically, these are different things, but you do it so that you can regard $N$ as a subset of $Z$. $\endgroup$
    – user49640
    Jun 13, 2017 at 14:49
  • $\begingroup$ @user49640 I see, so when I write $S_X = \langle X \rangle$ I'm considering $X$ as a set of one-symbol words that generate $S_X$? $\endgroup$ Jun 13, 2017 at 14:51
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    $\begingroup$ Right.${{{{{}}}}}$ $\endgroup$
    – user49640
    Jun 13, 2017 at 14:52

1 Answer 1

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As the comments explain, Milne is identifying $a\in X$ with $(a) \in S_X$ where I'm writing $(a)$ for a one element list/sequence/word. $(abc)$ would be a 3 element list. In category theory, there is a more general notion of subobject (and thus "subset") than the usual set-theoretic definition which is useful here. The notion of subobject instantiated for the category of sets and functions leads to a subobject of a set $Z$ being an injective function $\iota : W\to Z$. More precisely, it's an equivalence class of injective functions where $\iota_1 : W_1 \to Z$ and $\iota_2 : W_2 \to Z$ are equivalent if $\iota_1 = \iota_2\circ\varphi$ for some bijection $\varphi$. In fact, we can define a preorder on subobjects by saying $\iota_1\leq\iota_2$ when $\varphi$ is just an injection. Of course, every subset $Z'\subseteq Z$ in the usual sense determines the subobject via the inclusion $Z' \hookrightarrow Z$, and every injective function determines a subset $\iota[W]\subseteq Z$ via the direct image.

The upshot of all this is we can restate what Milne is saying without identifying anything. We're saying the map $\iota : X \to S_X$ defined as $\iota(x)=(x)$ is a subobject of $S_X$, and given any injective monoid homomorphism $f : M \to S_X$, if it "contains" $X$, meaning $\iota\leq f$, then $f$ is a bijection. This means we have some injective function $x \mapsto m_x : X \to M$ such that $f(m_x) = (x)$ at which point any element e.g. $(abc)\in S_X$ can be reconstructed as $f(m_a m_b m_c)$.

The benefit of this approach — besides getting familiar with some categorical concepts that will likely become increasingly useful and relevant to you — is that it doesn't require orchestrating a series of coincidences nor do you need to worry that something is true by coincidence. It is also (superficially in this case) more general. This approach doesn't really "cost" any more than Milne's, but it does shift the costs.

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  • $\begingroup$ Thanks a lot for this, I've been using Serge Lang alongside Milne's notes as a reference text and I expect the categorical concepts will come in handy $\endgroup$ Jun 14, 2017 at 11:41

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