Rotman's book An Introduction to the Theory of Groups (Fourth Edition) asks, on page 22, Exercise 2.8, to show that $S(n)$ cannot be embedded in $A(n+1)$, where $S(n)$ = the symmetric group on $n$ elements, and $A(n)$ = the alternating group on $n$ elements. I have a proof but it uses Bertrand's Postulate, which seems a bit much for page 22 of an introductory text. Does anyone have a more appropriate (i.e., easier) proof?
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6$\begingroup$ There is no such embedding for even $n$, just consider the orders of the respective groups: you don't have $|S{n}|$ dividing $|A{n+1}|$ by comparing the order of exponents of $2$. $\endgroup$– Olivier BégassatCommented May 15, 2011 at 21:55
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4$\begingroup$ By the way, you got your title wrong! $\endgroup$– Olivier BégassatCommented May 15, 2011 at 22:10
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2$\begingroup$ @Arturo: No, that's not really the rules of this forum. MathOverflow is mainly geared towards research-level mathematicians, and so any proof is fair game. That said, what you've highlighted is that this question probably isn't appropriate for MO; it could easily be closed as "too localized", which is our closest approximation to "homework-level". I would rather Len just accept Darij's answer below. Conversely, Len already says he has a proof. $\endgroup$– Theo Johnson-FreydCommented May 15, 2011 at 23:05
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3$\begingroup$ @Theo: I should have said "is not the intended answer by Rotman" rather than "cannot be used". $\endgroup$– Arturo MagidinCommented May 15, 2011 at 23:07
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2$\begingroup$ Someone should add the restriction of n > 1. Gerhard "Ask Me About System Design" Paseman, 2011.05.15 $\endgroup$– Gerhard PasemanCommented May 16, 2011 at 6:42
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6 Answers
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I think the following is sufficiently elementary: a transposition in $S_n$ is an element of order 2 commuting with at least $2(n-2)!$ elements of the group. But $A_{n+1}$ does not have such an element if $n$ is large enough. Indeed, if $\sigma\in A_{n+1}$ is of order 2, then it is a product of $k$ independent transpositions where $k$ is even and $2\le k\le(n+1)/2$. The number of elements of $A_{n+1}$ commuting with such $\sigma$ equals $2^{k-1}k!(n+1-2k)!$, and this is smaller than $2(n-2)!$ provided that $n\ge 6$.
answered May 16, 2011 at 17:42
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$\begingroup$ I am getting that $\sigma$ commutes with $2^{k-1}k!\frac{(n+1-2k)!}{2}$ elements. How did you arrive at that number? $\endgroup$ Commented Jul 20, 2017 at 14:51
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I think this is solved on http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=333049 .
answered May 15, 2011 at 21:51
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$\begingroup$ Thanks, that's a perfectly fine proof. And yet ... it uses concepts that Rotman has not yet introduced at page 22. $\endgroup$ Commented May 15, 2011 at 22:48
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1$\begingroup$ @Len: then you might want to be more precise about what concepts Rotman has introduced by page 22. $\endgroup$ Commented May 16, 2011 at 4:39
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3$\begingroup$ I wouldn't be surprised if this is just another mistake in Rotman. $\endgroup$ Commented May 16, 2011 at 9:31
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Of course this is not a research level question, and so is not appropriate for MO, but I remember being puzzled myself about what proof Rotman had in mind for this. I think we had better assume Lagrange's Theorem or it will be completely hopeless! Perhaps the proof using Bertrand's Postulate was intended, because students might expect to have heard of that, even if they have not read a proof?
Let's spell that out. As already noted, we can assume $n+1 = 2m$ is even by Lagrange. If $S_n$ embeds into $A_{n+1}$, then the index of the image of the embedding is $m$, so there is a nontrivial homomorphism (multiplicative action on cosets) $\phi: A_{n+1} \rightarrow S_m$.
By BP, there is a prime $p$ with $m < p < n+1$, so $p$ does not divide $|S_m|$. Hence all elements of order $p$ lie in ${\rm Ker}(\phi)$, including $g = (1,2,\ldots,p-1,p)$ and $h = (1,2,\ldots,p-1,p+1)$. Then $g^{-1}h$ is a 3-cycle ( $(1,p,p+1)$ if you multiply permutations left to right), so ${\rm Ker}(\phi)$ contains all 3-cycles, which generate $A_{n+1}$, contradicting the nontriviality of $\phi$.
answered May 16, 2011 at 10:10
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1$\begingroup$ You don't need Betran's Postulat for that argument: Small n can be done by hand and for $n\geq 4$ the group $A_{n+1}$ is simple, so that the homomorphism $A_{n+1}\to S_m$ must be injective. Now $m! < \frac{(n+1)!}{2}$ and that's the contradiction. $\endgroup$ Commented May 16, 2011 at 10:27
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4$\begingroup$ Yes but the whole point was to avoid using the simplicity of $A_{n+1}$, which has certainly not been covered by Page 22 of Rotman's book. $\endgroup$ Commented May 16, 2011 at 10:34
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$\begingroup$ That $A_{n+1}$ is generated by the 3-cycles - has that been established by page 22? $\endgroup$ Commented May 16, 2011 at 12:29
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2$\begingroup$ @Gerry: Actually, that is the immediately previous exercise to the one in question. (The one Len is asking about is 2.8; proving that $A_n$ is generated by 3-cycles for $n\gt 2$ is Exercise 2.7). $\endgroup$ Commented May 16, 2011 at 16:31
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$\begingroup$ @Derek: For that matter, cosets are not defined until page 24, right after this set of exercises, and Lagrange appears in page 26. The immediately previous exercise consists of showing that the 3-cycles generate $A_n$ for $n\gt 2$, so perhaps the argument Rotman has in mind consists of showing somehow that any putative image of $S_n$ in $A_{n+1}$ would necessarily contain all $3$-cycles. $\endgroup$ Commented May 16, 2011 at 16:53
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One could ask Rotman. It may be that in a reorganization of the material in the book that problem ended up earlier than the material needed for the (intended) answer. On the other hand it is not a bad experience for students to see problems where the complete solution seems slightly out of reach. Here, one can prove several small cases and see various potential directions for a general proof. Which will work? which are in the spirit of the subject? Of course it is best to set up the expectation that there might be problems like this.
answered May 16, 2011 at 18:13
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1$\begingroup$ I'd just like to add that it's true enough that this question is not quite appropriate for MO; but still, judging from the number of views it has received, even serious mathematicians like to relax once in a while with an elementary problem! $\endgroup$ Commented May 18, 2011 at 19:35
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Here is a short proof that follows from Rotman's material: Automorphisms of $S_m$ are all inner unless $m=2 \text{ or } 6$. It is not hard to use this to show that for $m\neq 2, 6$ subgroups of $S_m$ of index $m$ are in the form of $\{\sigma\mid \sigma(i)=i\}$ where $i\in\{1,\dots,m\}$ (see this stackexchange post).
Now a subgroup $H$ of $A_{n+1}$ isomorphic to $S_n$ yields an index-$(n+1)$ subgroup of $S_{n+1}$. Hence, except for $n=5$, $H$ must coincide with a stabilizer of the action $S_{n+1}\curvearrowright\{1,\dots,n+1\}$. But all such subgroups contain odd permutations, a contradiction. It remains to show that $S_5$ cannot be embedded in $A_6$. For this, notice that the former has elements of order six while the latter doesn't.
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Using concepts that Rotman have introduced before that exercise, I think we could provide the following counterexample: if $S_2$ could be embedded in $A_3$, then in $A_3$ there would be a subgroup $H$ with two elements. Such subgroup must contain $1$, so it must contain only one among $(1\ 2\ 3)$ and $(1\ 3\ 2)$; but if it contains one of them, it must contains the other one, as a power of the first element. So $A_3$ can not contain any subgroup with two elements, and $S_2$ can not be embedded in $A_3$.
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2$\begingroup$ The problem is that you only attempt to prove that $S_2$ cannot be embedded in $A_3$, but maybe the embedding is true for $n \ge 3$. $\endgroup$– Alex M.Commented Aug 9, 2022 at 12:15
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$\begingroup$ Sure, in Rotman's exercise the question was posed for $n\geq 2$, so the counterexample with $n=2$ was sufficient. $\endgroup$– DarioCommented Aug 9, 2022 at 13:37
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1$\begingroup$ Not quite: the problem could have easily been restated for $n \ge 3$. The point here is not to solve a problem, but to understand the deep reason for which a mathematical fact is true, which your solution does not do. In other words, your solution chooses to go for the "low-hanging fruit". $\endgroup$– Alex M.Commented Aug 9, 2022 at 13:48
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$\begingroup$ I think that this answers "does $S_n$ embed in $A_{n + 1}$ for all $n \ge 2$?" (in the negative), but it fails to prove "$S_n$ does not embed in $A_{n + 1}$ for all $n \ge 2$", of which the usual parenthesisation is "($S_n$ does not embed in $A_{n + 1}$) for all $n \ge 2$". $\endgroup$– LSpiceCommented Aug 9, 2022 at 15:27