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title: 'The alternating groups on more than four letters are simple',
clickbait: 'The alternating groups are the most accessible examples of simple groups, and this fact also tells us that the symmetric groups are "complicated" in some sense.',
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text: 'Let $n > 4$ be a [-45h]. Then the [-4hf] $A_n$ on $n$ elements is [simple_group simple].\n\n# Proof\n\nWe go by [mathematical_induction induction] on $n$.\nThe base case of $n=5$ [4jf we treat separately].\n\nFor the inductive step: let $n \\geq 6$, and suppose $H$ is a nontrivial [-4h6] of $A_n$.\nUltimately we aim to show that $H$ contains the $3$-[49f cycle] $(123)$; since $H$ [subgroup_normal_iff_union_of_conjugacy_classes is a union of conjugacy classes], and since the $3$-cycles [splitting_conjugacy_classes_in_alternating_group form a conjugacy class] in $A_n$, this means every $3$-cycle is in $H$ and [4hs hence] $H = A_n$.\n\n## Lemma: $H$ contains a member of $A_{n-1}$\nTo start, we will show that at least $H$ contains *something* from $A_{n-1}$ (which we view as all the elements of $A_n$ which don't change the letter $n$). %%note:Recall that $A_n$ [3t9 acts] naturally on the set of "letters" $\\{1,2,\\dots, n \\}$ by permutation.%%\n(This approach has to be useful for an induction proof, because we need some way of introducing the simplicity of $A_{n-1}$.)\nThat is, we will show that there is some $\\sigma \\in H$ with $\\sigma \\not = e$, such that $\\sigma(n) = n$.\n\nLet $\\sigma \\in H$, where $\\sigma$ is not the identity.\n$\\sigma$ certainly sends $n$ somewhere; say $\\sigma(n) = i$, where $i \\not = n$ (since if it is, we're done immediately).\n\nThen if we can find some $\\sigma' \\in H$, not equal to $\\sigma$, such that $\\sigma'(n) = i$, we are done: $\\sigma^{-1} \\sigma'(n) = n$.\n\n$\\sigma$ must move something other than $i$ (that is, there must be $j \\not = i$ such that $\\sigma(j) \\not = j$), because if it did not, it would be the transposition $(n i)$, which is not in $A_n$ because it is an odd number of transpositions.\nHence $\\sigma(j) \\not = j$ and $j \\not = i$; also $j \\not = n$ because if it were, [todo: this]\n\nNow, since $n \\geq 6$, we can pick $x, y$ distinct from $n, i, j, \\sigma(j)$.\nThen set $\\sigma' = (jxy) \\sigma (jxy)^{-1}$, which must lie in $H$ because $H$ is closed under [4gk conjugation].\n\nThen $\\sigma'(n) = i$; and $\\sigma' \\not = \\sigma$, because $\\sigma'(j) = \\sigma(y)$ which is not equal to $\\sigma(j)$ (since $y \\not = j$).\nHence $\\sigma'$ and $\\sigma$ have different effects on $j$ so they are not equal.\n\n## Lemma: $H$ contains all of $A_{n-1}$\nNow that we have shown $H$ contains some member of $A_{n-1}$.\nBut $H \\cap A_{n-1}$ is normal in $A_n$, because $H$ is normal in $A_n$ and it is [4h6 _intersect _subgroup _is _normal the intersection of a subgroup with a normal subgroup].\nTherefore by the inductive hypothesis, $H \\cap A_{n-1}$ is either the trivial group or is $A_{n-1}$ itself.\n\nBut $H \\cap A_{n-1}$ is certainly not trivial, because our previous lemma gave us a non-identity element in it; so $H$ must actually contain $A_{n-1}$.\n\n## Conclusion\n\nFinally, $H$ contains $A_{n-1}$ so it contains $(123)$ in particular; so we are done by the initial discussion.\n\n# Behaviour for $n \\leq 4$\n- $A_1$ is the trivial group so is vacuously not simple.\n- $A_2$ is also the trivial group. \n- $A_3$ is isomorphic to the [-47y] $C_3$ on three generators, so it is simple: it has no nontrivial proper subgroups, let alone normal ones.\n- $A_4$ has the following normal subgroup (the [klein_four_group Klein four-group]): $\\{ e, (12)(34), (13)(24), (14)(23) \\}$. Therefore $A_4$ is not simple.',
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