3
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I used the fact that $A_5$ is the only real normal subgroup of $S_5$. Then we have a 2 types of homomorphisms and the trivial homomorphism.

  • homomorphisms with a kernel $\{e\}$

  • homomorphisms with a kernel $A_5$

In the first case the first isomorphism theorem implies that the image of the homomorphism is isomorphic to $S_5$, therefore the number of homomorphism of that type is the number of injections of $S_5$ in $A_6$.

In the second case because the index of $A_5$ in $S_5$ is 2, the quotient group is isomorphic to $C_2$. Therefore the image of the homomorphism is isomorphic to $C_2$.

That's as far as I got, I classified the homomorphisms but I don't know how to count them.

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5
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There is no injective homomorphism from $S_5$ to $A_6$. Its image would have index $3$ in $A_6$. There is no such subgroup of $A_6$. If there were, there would be a homomorphism from $A_6$ onto a group of order $3$ or $6$, and there isn't.

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-2
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This answer is wrong, I left it undeleted to remind me of my fault, please notice this and don't be misguided.

Another reason for there is no injective homomorphism from S5 to A6 is that, if so S5 can be seen as a subgroup of A6 but every element in A6 is even permutation while S5 has odd permutation.

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  • 3
    $\begingroup$ If this argument were correct, it would also show that there's no injective homomorphism from $S_5$ to $A_7$. But there is such a homomorphism: Send every even permutation of $\{1,2,3,4,5\}$ to itself, leaving $6$ and $7$ fixed, and send every odd permutation of $\{1,2,3,4,5\}$ to its product with the transposition $(67)$. $\endgroup$ – Andreas Blass Feb 7 '18 at 19:29
  • $\begingroup$ Thank you for correct my fault $\endgroup$ – Bingyan Liu Feb 8 '18 at 3:14

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