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Prove or disprove that there is an onto homomorphism from $S_4\to S_3$ where $S_n$ is the symetric group of order $n!$.

after long time of searching, I finally success but i just manually tried to associated to element of $S_4$ an element of $S_3$. I wonder if there is an easier technic.

user349449
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1 Answers1

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Hint

Let $V=\{1,(12)(34),(13)(24),(14)(23)\}$ a subgroup of $\mathfrak S_4$. This group has 4 element, and is normal in $\mathfrak S_4$. If you can prove that $$\mathfrak S_4/V\cong \mathfrak S_3,$$ then, if $$\varphi:\mathfrak S_4/V\longrightarrow \mathfrak S_3$$ is such an isomorphism, then, $\varphi\circ \pi$ is the researched homomorphism where $$\pi:\mathfrak S_4\longrightarrow \mathfrak S_4/V$$ is the canonical surjection.

Added

You don't really have to construct it. $\mathfrak S_4/V$ is a group with $6$ elements. Therefore it's isomorphic either to $\mathfrak S_3$ or to $\mathbb Z/6\mathbb Z$. For example, in $\mathfrak S_4/V$, $(12)$ and $(23)$ don't commute, therefore it's not $\mathbb Z/6\mathbb Z$, and thus, it must be $\mathfrak S_3$.

Surb
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  • @joriki: I corrected it, but even $\left<(12),(34)\right>$ doesn't work since $(12)\notin V$ and $(13)(24)$ can't be written as a combinaison of $(12)$ and $(34)$. But of course, $\left<(12),(34)\right>$ and $V$ are isomorphic. – Surb Jun 26 '16 at 10:48
  • I don't see how to construct the isomorphism, could you give me a hint ? – user349449 Jun 26 '16 at 11:42