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Claim:

Short exact sequence $1 \longrightarrow \operatorname{Inn}\left(A_{6}\right) \longrightarrow \operatorname{Aut}\left(A_{6}\right) \longrightarrow \operatorname{Out}\left(A_{6}\right) \longrightarrow 1$ is not right split,

where $\operatorname{Inn}\left(A_{6}\right)\cong A_6$, $\operatorname{Aut}\left(A_{6}\right)\cong\operatorname{Aut}\left(S_{6}\right)\cong S_6\rtimes \mathbb Z_2$ and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2$.

To move this question out of the unanswered list, I put my proof in the answer.

Thanks for your time and patience :)

Derek Holt
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Andrews
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    The proof is basically correct (congratulations on that!), but you haven't justified your claim that $\rho$ and $\sigma$ can both be considered as elements of ${\rm Aut}(S_6) \setminus {\rm Inn}(S_6)$, and in fact it is not necessarily true. But at least one of them, say $\rho$, must lie outside of ${\rm Inn}(S_6)$, and if $\sigma \in {\rm Inn}(S_6)$, then we can replace $\sigma$ by $\sigma \rho$, and everything works. – Derek Holt Nov 01 '19 at 12:18
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    @DerekHolt I fixed it. Dear Derek Holt, I remember you since you helped me so many times. Thank you very much!! – Andrews Nov 01 '19 at 14:06

1 Answers1

1

Claim:

Short exact sequence $1 \longrightarrow \operatorname{Inn}\left(A_{6}\right) \longrightarrow \operatorname{Aut}\left(A_{6}\right) \longrightarrow \operatorname{Out}\left(A_{6}\right) \longrightarrow 1$ is not right split,

where $\operatorname{Inn}\left(A_{6}\right)\cong A_6$, $\operatorname{Aut}\left(A_{6}\right)\cong\operatorname{Aut}\left(S_{6}\right)\cong S_6\rtimes \mathbb Z_2$ and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2$.

Proof:

$1$. Prerequisites:

(1) Element in $\operatorname{Aut}\left(A_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$ swaps conjugate classes $(abc)$ and $(abc)(def)$ in $A_6$.

(2) Element in $\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ swaps conjugate classes $(ab)$ and $(ab)(cd)(ef)$ in $S_6$,

swaps conjugate classes $(abc)$ and $(abc)(def)$ in $A_6$(also in $S_6$).

$2$. Suppose the sequence right splits and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\leqslant\operatorname{Aut}\left(A_{6}\right)$

where $\sigma, \rho\in \operatorname{Aut}\left(A_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\langle \sigma\rangle\langle\rho\rangle\cap\operatorname{Inn}(A_6)=1$.

Since $\operatorname{Aut}\left(A_{6}\right)=\operatorname{Aut}\left(S_{6}\right)$, $\rho$ and $\sigma$ can be considered as elements in $\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$.

If $\sigma, \rho\in \operatorname{Inn}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\sigma\rho\in \operatorname{Inn}\left(A_{6}\right)$. Contradiction.

If $\rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ and $\sigma\in \operatorname{Inn}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\sigma\rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.

$\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\cong\langle \sigma\rho\rangle\langle\rho\rangle$ and $\sigma\rho, \rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.

So if the sequence right splits, we can always assume

$\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\leqslant\operatorname{Aut}\left(A_{6}\right)$ where $\rho, \sigma\in\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.

$3$. $[\operatorname{Aut}\left(S_{6}\right):\operatorname{Inn}\left(S_{6}\right)]=2$ , $\sigma\operatorname{Inn}\left(S_{6}\right)=\rho\operatorname{Inn}\left(S_{6}\right)$, $\rho^{-1}\sigma\in\operatorname{Inn}(S_6)$.

Suppose $\rho^{-1}\sigma=c_\gamma$, where $c_\gamma$ is action of conjugation by $\gamma\in S_6$.

Since $\langle \sigma\rangle\langle\rho\rangle\cap\operatorname{Inn}(A_6)=1$, $\gamma\in S_6\setminus A_6$ is odd permutation.

$(\rho^{-1}\sigma)^2=c_\gamma^2=1$ gives $\gamma^2=1$, $\gamma$ is transposition or product of three disjoint transpositions.

$\sigma\rho=\rho\sigma$ gives $\rho(\gamma)=\gamma$.

But $\rho\in\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ swaps conjugate classes $(ab)$ and $(ab)(cd)(ef)$.

Contradiction. $\Box$

Andrews
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