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text: '[summary: The [-5kg] may be constructed formally using van der Waerden's trick, which is not intuitive at all but leads to a definition that is very easy to work with. This page will detail van der Waerden's construction, and will prove that the trick yields a group which has all the right properties to be the free group.] \n\nThe [-5kg] may be constructed formally using van der Waerden's trick, which is not intuitive at all but leads to a definition that is very easy to work with. This page will detail van der Waerden's construction, and will prove that the trick yields a group which has all the right properties to be the free group.\n\n# The construction\n\nWrite $X^r$ for the set that contains all the [5jc freely reduced words] over $X \\cup X^{-1}$ (so, for instance, excluding the word $aa^{-1}$). %%note:We use the superscript $r$ to denote "reduced".%%\n\nWe define the *free group* $F(X)$, or $FX$, on the set $X$ to be a certain [-576] of the [-497] $\\mathrm{Sym}(X^r)$: namely that which is generated by the following elements, one for each $x \\in X \\cup X^{-1}$:\n\n- $\\rho_x : \\mathrm{Sym}(X^r) \\to \\mathrm{Sym}(X^r)$, sending $a_1 a_2 \\dots a_n \\mapsto a_1 a_2 \\dots a_n x$ if $a_n \\not = x^{-1}$, and $a_1 a_2 \\dots a_{n-1} x^{-1} \\mapsto a_1 a_2 \\dots a_{n-1}$.\n- $\\rho_{x^{-1}} : \\mathrm{Sym}(X^r) \\to \\mathrm{Sym}(X^r)$, sending $a_1 a_2 \\dots a_n \\mapsto a_1 a_2 \\dots a_n x^{-1}$ if $a_n \\not = x$, and $a_1 a_2 \\dots a_{n-1} x \\mapsto a_1 a_2 \\dots a_{n-1}$.\n\nRecall that each $\\rho_x$ lies in $\\mathrm{Sym}(X^r)$, so each is a [-499] from $X^r$ to $X^r$.\nWe specify it by stating what it does to every element of $X^r$ (that is, to every freely reduced word over $X$).\n\nWe first specify what it does to those words which don't end in $x^{-1}$: $\\rho_x$ simply appends an $x$ to such words.\nWe then specify it to the remaining words, those which do end in $x^{-1}$: then $\\rho_x$ just removes the $x^{-1}$.\n\nIt's easy to check that if $\\rho_x$ is given a freely-reduced word as input, then it produces a freely-reduced word as output, because the only change to the word is at the end and we make sure to provide a separate definition if $x$ is to be cancelled.\nTherefore each $\\rho_x$ is a function $X^r \\to X^r$.\n\nThen we do it all again for all the inverses $x^{-1}$, creating the functions $\\rho_{x^{-1}}$; and finally, we add in the [-54p], denoted $\\rho_{\\varepsilon}$, which simply returns its input unchanged.\n\nNotice that the $\\rho_x$ and $\\rho_{x^{-1}}$ are all indeed bijective (and therefore members of $\\mathrm{Sym}(X^r)$), because in fact $\\rho_x$ and $\\rho_{x^{-1}}$ are [4sn inverse] to each other (each cancelling off what the other did), and a function with an inverse is bijective.\n\nSo, we've defined the free group as a certain subgroup of the symmetric group.\nRemember that the subgroup has as its group operation "function composition"; so $\\rho_x \\cdot \\rho_y = \\rho_x \\circ \\rho_y$, for instance.\nWe will write $\\rho_x \\rho_y$ for this, omitting the group operation.\n\nSomething key to notice is that if we apply $\\rho_{a_n} \\rho_{a_{n-1}} \\dots \\rho_{a_1}$ to the empty word $\\varepsilon$, we get $$\\rho_{a_n} \\rho_{a_{n-1}} \\dots \\rho_{a_1}(\\varepsilon) = \\rho_{a_n} \\rho_{a_{n-1}} \\dots \\rho_{a_3}(\\rho_{a_2}(a_1)) = \\rho_{a_n a_{n-1} \\dots a_3}(a_1 a_2) = \\dots = a_1 a_2 \\dots a_n$$\nif $a_1 a_2 \\dots a_n$ is a freely reduced word.\n(Indeed, if the word is freely reduced then none of the successive $\\rho_{a_i}, \\rho_{a_{i+1}}$ can have cancelled each other's effect out, so every application of a $\\rho_{a_i}$ must be appending a letter.)\nHence we might hope to have captured the freely reduced words in our subgroup.\n\n# The formal definition is the same as the intuitive definition\n\nWe'll show that there is a bijection between the free group and the set of reduced words, by "converting" each reduced word into a corresponding member of the free group.\n\nTake a reduced word, $w = a_1 a_2 \\dots a_n$, and produce the member of the free group (that is, the function) $\\rho_{a_1} \\rho_{a_2} \\dots \\rho_{a_n}$. %%note:Recall that the group operation here is composition of functions, so this is actually the function $\\rho_{a_1} \\circ \\rho_{a_2} \\circ \\dots \\circ \\rho_{a_n}$.%%\nThis really does produce a member of the free group (i.e. of the subgroup of the symmetry group), because each $a_i$ is an element of $X \\cup X^{-1}$ and we have already specified how to make $\\rho_{a_i}$ from such an element.\n\nNow, we claim that in fact this map is injective: that is, we can't take two words $a_1 a_2 \\dots a_n$ and $b_1 b_2 \\dots b_m$ and produce the same member of the free group.\n(That is, we show that $\\rho_{a_1} \\rho_{a_2} \\dots \\rho_{a_n} = \\rho_{b_1} \\rho_{b_2} \\dots \\rho_{b_m}$ implies $a_1 \\dots a_n = b_1 \\dots b_m$.)\nIndeed, if the two functions ("elements of the free group") are equal, then they must in particular do the same thing when they are applied to the empty word $\\varepsilon$.\nBut by the "key notice" above, when we evaluate $\\rho_{a_1} \\rho_{a_2} \\dots \\rho_{a_n}$ at the empty word, we get $a_n a_{n-1} \\dots a_2 a_1$; and when we evaluate $\\rho_{b_1} \\rho_{b_2} \\dots \\rho_{b_m}$ at the empty word, we get $b_m b_{m-1} \\dots b_2 b_1$; so the two words must be equal after all. [todo: we've got this backwards]\n\nFinally, the map is surjective: we can make any member of the free group by "converting the appropriate reduced word into a function".\nIndeed, the free group is generated by the $\\rho_x$ and $\\rho_{x^{-1}}$ for $x \\in X$, so every element is some $\\rho_{x_1} \\dots \\rho_{x_n}$ for some selection of $x_1, \\dots, x_n \\in X \\cup X^{-1}$.\nNote that $x_1 \\dots x_n$ need not necessarily be freely reduced as a word at the moment; but if it is indeed not freely reduced, so some $x_i, x_{i+1}$ cancel each other out, then removing that pair completely doesn't change the function $\\rho_{x_1} \\dots \\rho_{x_n}$.\nFor example, $\\rho_{x_1} \\rho_{x_1^{-1}} \\rho_{x_2} = \\rho_{x_2}$.\nHence the process of "performing one step of a free reduction" (i.e. removing a cancelling pair) doesn't change the member of the free group as a function; and since each such removal makes the word shorter, it must eventually terminate.\nIt remains to show that it doesn't matter in what order we remove the cancelling pairs; but that is immediate because we've already shown that our "conversion" process is injective: we started with a member of the free group, so if it corresponds to a freely reduced word then it corresponds to a *unique* freely reduced word.\nSince we've just shown that it does indeed correspond to a freely reduced word (by repeatedly removing cancelling pairs), we are done.\n\nThe above shows that the free group can be considered just to be the set of reduced words.',
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