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clickbait: 'An easy way to determine that many groups are not free: free groups contain no non-identity elements of finite order.',
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text: 'Given the [-5kg] $FX$ on a set $X$, it is the case that $FX$ has no [torsion_group_theory torsion] elements.\nThat is, every element has infinite order except for the [-54p].\n\n# Proof\n\nRecall that we can view every element of the free group as being a [-5jc] whose letters are elements of $X$.\nAlso the group operation is "concatenation of words, followed by free reduction".\nIt is certainly intuitively plausible that if we repeatedly concatenate a word with more copies of itself, and then perform free reduction, we will never reach the empty word; this is what we shall prove.\nThe cancellations in the process of free reduction are everything here, because the only way the powers of a word can get shorter (and hence the only way the powers of a word can ever come to the empty word) is by those cancellations.\nSo we are going to have to analyse the behaviour of words at their start and ends, because when we take powers of a word, the start and end are the places where cancellation may happen.\n\nWe will say a word is *cyclically reduced* if it is not only freely reduced, but also it is "freely reduced if we rotate the word round one place".\nFormally, a freely reduced word $a_1 a_2 \\dots a_n$ is cyclically reduced if and only if $a_1 \\not = a_n^{-1}$.\nThis captures the idea that "the word doesn't admit any cancellation when we take powers of it".\n\n%%hidden(Examples):\n[todo: some examples of words which are cyclically reduced and some which are not]\n%%\n\nNow, every freely reduced word may $w$ be written as $r w^\\prime r^{-1}$ where $r$ is a freely reduced word and $w^\\prime$ is a cyclically reduced word, and there is no cancellation between $r$, $r^{-1}$ and $w^\\prime$.\nThis is easily proved by [-5fz] on the length of $w$: it is immediate if $w$ is already cyclically reduced (letting $r = \\varepsilon$ the empty word, and $w^\\prime = w$), while if $w$ is not cyclically reduced then it is $a v a^{-1}$ for some letter $a \\in X$ and some freely reduced word $v$.\nBut then by the inductive hypothesis (since $v$ is shorter than $w$), $v$ may be written as $r v^\\prime r^{-1}$ where $v^\\prime$ is cyclically reduced; so $w = a r v^\\prime r^{-1} a^{-1} = (ar) v^\\prime (ar)^{-1}$ as required.\n\nMoreover, this decomposition is *unique*, since if $r w^\\prime r^{-1} = s v^\\prime s^{-1}$ then $s^{-1} r w^\\prime r^{-1} s = v^\\prime$; but $v^\\prime$ is cyclically reduced so there are only two ways to prevent cancellation:\n\n- $s$ is the identity, whereupon $v^\\prime = r w^\\prime r^{-1}$ is cyclically reduced, so (by freely-reducedness of $w = r w^\\prime r^{-1}$) we have $r = e$ as well, and $v^\\prime = w^\\prime = w$.\n- $s$ is not the identity but it entirely cancels with something in $r^{-1}$; hence $s$ is a sub-word of $r$. But by symmetry, $r$ is a sub-word of $s$, so they are the same (because without loss of generality $r$ is also not the identity); and therefore $v^\\prime = w^\\prime$.\n\nFinally, to complete the proof that the free group is torsion-free, simply take a putative word $w$ whose order $n$ is finite.\nExpress it uniquely as $r w^\\prime r^{-1}$ as above; then $(rw^\\prime r^{-1})^n = r (w^\\prime)^n r^{-1}$ which is expected to become the empty word after free reduction.\nBut we already know there is no cancellation between $r$ and $w^\\prime$ and $r^{-1}$, so there cannot be any cancellation between $r, (w^\\prime)^n, r^{-1}$ either, and by the cyclically-reduced nature of $w^\\prime$, our power $r (w^\\prime)^n r^{-1}$ is actually freely reduced already.\nHence our power of the word is emphatically not the empty word.',
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