Non-existence of $\mathfrak{S}_n \hookrightarrow \mathfrak{A}_{n+1}$.

Asked Apr 28 '15 at 18:48

Active Apr 28 '15 at 18:54

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Let $\mathfrak{S}_n$ be the symmetric group (permutations of $n$ items) and let $\mathfrak{A}_n$ be the alternate group.

For $n \geq 5$, I have to show that there is no injective morphism $\mathfrak{S}_n \hookrightarrow \mathfrak{A}_{n+1}$.

If such a morphism exists, then its image $H$ is a subgroup of order $n!$ of $\mathfrak{A}_{n+1}$ which is of order $(n+1)!/2$, so $H$ has index $(n+1)/2$ and this is only possible when $n$ is odd. That's all I have... I guess the simplicity of $\mathfrak{A}_{n+1}$ will be a key argument but I don't see where I could use it.

Hints and/or solutions are welcome !

edited Apr 28 '15 at 18:54

asked Apr 28 '15 at 18:48

Tlön Uqbar Orbis Tertius

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You should probably say if $\mathfrak{S}_n$ and $\mathfrak{A}_n$ are the symmetric and alternating groups on $n$ objects: the fraktur font is not the usual notation. – Chappers Apr 28 '15 at 18:51
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Ah, in my country (France), it is undoubtingly the most common notation. I'll specify it. – Tlön Uqbar Orbis Tertius Apr 28 '15 at 18:52
If I remember the strategy correctly, you should look for some element of $S_n$ which has too big centralizer for $A_{n+1}$ being able to provide. – Pavel Čoupek Apr 28 '15 at 18:57
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Also, it is a duplicate of this and the answer can be found (for some reason) here. – Pavel Čoupek Apr 28 '15 at 19:00
Thank you, I didn't see this topic. – Tlön Uqbar Orbis Tertius Apr 28 '15 at 19:01

Non-existence of $\mathfrak{S}_n \hookrightarrow \mathfrak{A}_{n+1}$.

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