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It is well known that the non abelian exterior square of $S_n$ is again a group of order $n!~(n>2)$ e.g for $n=5, ~ S_n\wedge S_n\cong \text{SL}(2,5)$ (https://groupprops.subwiki.org/wiki/Exterior_square_of_a_group). My question is that:

Is this possible that for some $n$, $S_n\wedge S_n\cong S_n$? I have seen that it is not possible for $n<7$, but above $7$, I do not have any information.

Please help me to solve this problem.

MANI
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1 Answers1

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Looking at the link you provided, there is a surjective morphism $S_n\wedge S_n\to [S_n,S_n]=A_n$, so $A_n$ is a quotient of $S_n\wedge S_n$. But $A_n$ is not a quotient of $S_n$ since, for $n>2$, $S_n$ does not have a normal subgroup of order $2$.

Captain Lama
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  • Thanks for your answe sir, your logic is perfect and thanks for the answer. How can we see the surjective morphism? I think my problem has bee solved by seeing that $S_n \wedge S_n$ has a normal subgroup of order $2$ but $S_n$ does not have. But if you explain this that how $g\wedge h \mapsto [g,h]$ is onto. – MANI Nov 10 '19 at 19:29
  • The non-obvious part is that $g\wedge h\mapsto [g,h]$ is well-defined (by the universal property of the exterior square). The fact that it is surjective is easy, since $[G,G]$ is by definition generated by the commutators $[g,h]$, which are by construction in the image. – Captain Lama Nov 10 '19 at 19:39
  • I was also thinking the same but what for abelian groups? – MANI Nov 10 '19 at 19:40
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    Well, if $G$ is abelian then $[G,G]=0$, so $G\wedge G\to [G,G]$ is the zero morphism, which shows that the map $H_2(G,\mathbb{Z})\to G\wedge G$ is an isomorphism in that case. – Captain Lama Nov 10 '19 at 21:31