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Well, we know all normal subgroups of $S_n$, including the interesting case $n=4$. Using this one can conclude that the cyclic group of order 3 is not a quotient of $S_n$. Is there are more direct way to see this?

$C_3$ is simple, so if there were a surjective homomorphism from $S_n$ onto $C_3$, its kernel would be a maximal subgroup of index 3. The alternating group is also maximal as it has index 2. Can we combine these two fact together to get a contradiction?

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Here's an answer that does not require anything about generation of $S_n$, but only knowledge about conjugacy in $S_n$ (which is perhaps easier):

Let $f$ be a homomorphism from $S_n$ to $C_3$ and let $x$ be an element of $S_n$. Note that $x$ is conjugate to its inverse (since they have the same cycle structure). Since $f$ is a homomorphism, $f(x)$ is conjugate (in $C_3$) to its inverse. Since $C_3$ is abelian, $f(x)$ is equal to its inverse and thus $f(x)=1$.

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Let $G$ be a normal subgroup of index $3$ in $S_n$. Then $g^3\in G$ for all $g\in S_n$. In particular, every transposition $(i\, j)$ is inside $G$. Since the transpositions generate $S_n$, we see that $G = S_n$.

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    $\begingroup$ Beautiful answer. $\endgroup$ – Alex Wertheim Jan 25 '18 at 23:20
  • $\begingroup$ I love this.... $\endgroup$ – Randall Jan 25 '18 at 23:41
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Suppose $N \unlhd S_n$, and $|S_n:N|=3$. Then $n \geq 3$ and $S_n/N$ is abelian. Hence, $A_n=[S_n,S_n] \subseteq N$, which is absurd because $|S_n:A_n|=2$.

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