This question is about the (in)famous Rotman's exercise 2.8 in "An Introduction to the Theory of Groups." I've searched and found similar questions here and in MO, but none of them contains a valid proof. (Does $S_n$ belong as a subgroup to $A_{n+1}$?)

According to Rotman, a valid proof can only use the concepts introduced up to this exercise: cycle permutations, factorization of permutations, odd and even permutations, semigroups, groups, homomorphism and subgroups. Cosets, Lagrange's theorem, normal subgroups, and so on are not yet introduced. I stress this point because all of the proofs I've seen use Lagrange or actions, on cosets.

Now my attempt is to use exercise 2.7 (solved) which is about a proof that $A_n$ ($n>2$) is generated by all the $3$-cycles and exercise 2.4 (solved) " if $S$ is a proper subgroup of $G$ then $\langle G \setminus S\rangle=G$ " in this way:

Suppose that for every $\phi : S_n \to A_{n+1}$ imbeddings, all the $ 3$-cycles are contained in $\operatorname{Im}\phi $, then the assertion is proved by absurd. But I can't find a way to prove if it is possible or either find a counterexample to this kind of approach.

If someone has another proof which use only basics concepts is well accepted of course, but I mainly need some hints about correctness or not of my reasoning and how to proceed if it is correct. Thank you in advance.

  • $\begingroup$ I am not very sure about the way i am following as you said i can not use more group theory but, Suppose $S_n$ is in $A_{n+1}$, what is the order of $S_n$ and what is the order of $A_{n+1}$... does $n!$ divides $\frac{(n+1)!}{2}$ in general.... $\endgroup$
    – user87543
    Aug 18, 2013 at 14:02
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    $\begingroup$ Have you tried counting the number of elements of order $2$ in both $S_n$ and $A_{n+1}$? The computations are rather cumbersome, but it requires no knowledge of group theory. $\endgroup$
    – Servaes
    Aug 18, 2013 at 19:51
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    $\begingroup$ @user84559: I thought counting elements of order 2 might work too, but it doesn't. To save doing the combinatorics here, oeis.org/A000085 and oeis.org/A000704 give the relevant counts and the only case where $S_n$ has more elements of order 2 than $A_{n+1}$ is when $n = 2$. $\endgroup$
    – Rob Arthan
    Aug 19, 2013 at 16:29
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    $\begingroup$ Isn't there a result that the largest subgroup of $A_{n+1}$ is $A_n$? I may be mistaken. $\endgroup$
    – 2'5 9'2
    Aug 19, 2013 at 20:49
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    $\begingroup$ Here is a link to an answer at MO: mathoverflow.net/questions/65083/… $\endgroup$
    – 2'5 9'2
    Aug 19, 2013 at 20:57

1 Answer 1


I'm not sure what exactly I'm allowed to use or what Rotman had in mind, but I think this approach uses no more group theory than the (unique) factorization of permutations and the basic idea of what a homomorphism is.

If $n$ is odd there exists a subset $S\subset S_n$ of pairwise commuting elements of order $2$ with $|S|=\tfrac{n-1}{2}$. For example $$\{(1\ 2),(3\ 4),\ldots,(n-2\ n-1)\}.$$ An element $\sigma\in A_{n+1}$ of order $2$ is necessarily a product of an even number of disjoint $2$-cycles. Therefore a subset $A\subset A_{n+1}$ of pairwise commuting elements of order $2$ cannot contain more than $\tfrac{n+1}{4}$ elements. Hence for an embedding $S_n\ \hookrightarrow\ A_{n+1}$ to exist we must have $$\frac{n-1}{2}\leq\frac{n+1}{4},$$ or equivalently $n\leq3$. At this point it should not be hard to verify that $S_3$ does not embed into $A_4$, and of course that $S_1$ does embed into $A_2$.

A similar argument can be made when $n$ is even, to find that for an embedding $S_n\ \hookrightarrow\ A_{n+1}$ to exist we must have $$\frac{n}{2}<\frac{n+1}{4},$$ or equivalently $n<1$, immediately yielding a contradiction.

  • $\begingroup$ Woow, I need to sit down and check it, anyway, thanks for this answer, seems to be plausible, but again, I've yet to read it carefully! :) $\endgroup$
    – Riccardo
    Jul 1, 2015 at 21:19
  • $\begingroup$ Though it seemed like a good idea at 03:00 AM last night, I see now that my proof doesn't work. But perhaps it can be saved. I'll give it some more thought. $\endgroup$
    – Servaes
    Jul 1, 2015 at 21:48
  • $\begingroup$ From your comment I conclude that you realized in meanwhile that $A_4$ has an elementary abelian subgroup of order $4$, i.e., two commuting elements of order $2$. $\endgroup$
    – j.p.
    Jul 2, 2015 at 10:52

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