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I'm working through Rotman (1994). It's the 2nd chapter where my curiosity lead me to a question that is not from the book. Here's the problem that inspired the question:

2.8. Imbed $S_n$ as a subgroup of $A_{n+2}$, but show, for $n\ge2$, that $S_n$ cannot be imbedded in $A_{n+1}.$

Since the alternate subgroup has the property of even parity, I thought that a little investigation of parity for the second and third symmetric groups was in order. So I constructed a table for each of them, grouped by even and odd number of disjoint products.

I'll forego writing the actual table and simply provide the answers. In symmetric groups with an even idegree, two even permutations come together under composition to produce an even permutation. Two odds also make an even. Two different parities produce odd elements.

When the symmetric group has an odd degree, e.g. $S_3$, two odds make an odd, two evens make an odd, and two different parities make an even element. Exactly the opposite results from the symmetric groups with even index.

Two questions: is this enough to prove the $A_{n+1}$ case? And can this be generalized modulo 2 e.g. $A_{2n+1}$?

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    You aren't computing the parity correctly. The multiplication rule should be the same for all $S_n$. Also, $A_n$ embeds into $A_{n+1}$ for all $n$.2011-11-17
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    @bwkaplan: Why did you change back the title? The question asks that you embed $S_n$ into $A_{n+2}$, *not* into $A_{n+3}$. Your entire question never mentions $A_{n+3}$.2011-11-17
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    @ArturoMagidin, I am more interested in the $A_{2n+1}$ groups, with the first one e.g. $A_{n+1}$ being a special case. See the last paragraph. Anyway, I'm going off to recalculate the parity again and will be back if I have any more questions. thanks again!2011-11-17
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    @bwkaplan: I think only Qiaochu Yuan has read your (mind to be able to read your) actual question, but perhaps his answer was not direct enough. **Yes**, Sn is isomorphic to a subgroup of A(n+3). You answer the exercise to get Sn isomorphic to a subgroup of A(n+2), and then use QY's answer that A(n+2) is isomorphic to a subgroup of A(n+3).2011-11-17
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    @bwkaplan: Still, $2n+1\neq n+3$...2011-11-17
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    @ArturoMagidin, Fixed.2011-11-17

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