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Interesting problem from symmetric group:

For $ \ n \geq 3 \ $ , if $ \ \alpha \in S_n \ \ define \ \ \alpha' \in S_{n+2} \ $ by $ \begin{Bmatrix} \alpha'(i)=\alpha(i) \ \ 1 \leq i \leq n \\ \alpha'(n+1)=n+1 \ \ \ \ \ \ \ \ \ \\ \alpha'(n+2)=n+2 \ \ \ \ \ \ \ \ \end{Bmatrix} $

Define $ \phi : S_n \rightarrow S_{n+2} $ by

$ \ \ \phi(\alpha)=\alpha ' \ \ if \ \ \alpha \in A_n \ \ $ and

$ \phi(\alpha)=\alpha' \beta \ \ if \ \ \alpha' \notin A_n \ \ for \ \ \beta= (n+1 \ \ \ n+2) \ cycle $

(a) Prove that $ \ \phi \ $ is homomorphism .

(b) Prove that $ \ \ \phi(S_n) \leq A_{n+2} \ $.

My approach :

(a) Let $ a , \ b \in S_n $ . If $ a,b \in A_n , then \ \ ab \in A_n , \ \ because \ \ A_n \ \ forms \ a \ group . $

Then $ \ \phi(ab)(i)=(ab)'(i)=(ab)(i)=a(i) b(i)=a'(i) b'(i)=\phi(a) \phi(b)(i) $

or, $ \phi(ab)=\phi(a) \phi(b) $.

Hence $ \phi \ is \ a \ homomorphism . $

I think for the part (b) we have to show that $ \ \phi:S_n \rightarrow A_{n+2} $ is an Injection map . But I am unable to show.

Am I right ? Any help for the second part ?

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    $\begingroup$ You only proved it for the special case of when $a$ and $b$ are both in $A_n$. $\endgroup$ – Akiva Weinberger Jul 14 '17 at 3:28
  • $\begingroup$ I can not proceed further . Any help is really appreciating $\endgroup$ – M. A. SARKAR Jul 14 '17 at 3:29
  • $\begingroup$ What happens if $a\in A_n$ and $b\notin A_n$? $\endgroup$ – Akiva Weinberger Jul 14 '17 at 3:29
  • $\begingroup$ If $ a \in A_n \ and \ b \notin A_n $ , then $ ab \notin A_n $. So $ \phi (ab)=(ab)' \beta=(a' ) (b' \beta)=\phi(a) \phi(b) $ $\endgroup$ – M. A. SARKAR Jul 14 '17 at 3:56
  • $\begingroup$ What if $a\notin A_n$ and $b\in A_n$? $\endgroup$ – Akiva Weinberger Jul 14 '17 at 4:01
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a-)

To show that $\phi$ is an homomorphism, let $\alpha_1, \alpha_2 \in S_n$.

If $\alpha_1, \alpha_2 \in A_n$.Then since $\alpha_1 \circ \alpha_2 \in A_n$, $$\phi (\alpha_1 \circ \alpha_2) = (\alpha_1 \circ \alpha_2)^{'} = (\alpha_1)^{'} (\alpha_2)^{'}$$

If, without lost of generality, $\alpha_1 \not \in A_n$ and $\alpha_2 \in A_n$,

$$\phi (\alpha_1 \circ \alpha_2) = (\alpha_1 \circ \alpha_2)^{'} \circ \beta = (\alpha_1^{'}) (\alpha_2^{'}) \beta$$ since $(n+1)$ and $(n+2)$ are fix in $\alpha_2$, $$\phi (\alpha_1 \circ \alpha_2) = (\alpha_1^{'})\beta (\alpha_2^{'}$$

If $\alpha_1, \alpha_2 \not \in A_n$, then

$$\phi (\alpha_1 \circ \alpha_2) = (\alpha_1 \circ \alpha_2)^{'}= \alpha_1^{'} \circ \alpha_2^{'}$$ since $\alpha_1 \circ \alpha_2 \in A_n$.

b-) Note that in all of the if statements above, $\phi (...) \in A_n$, and for the rest, I think it is clear.

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To show part (b)(i.e., $\phi(S_n)\subset A_{n+2}$), I think it is enough to show that the image of the function $\phi$ lies in $A_{n+2}$ i.e., to show that $\phi(\alpha)\in A_{n+2}$ for all $\alpha \in S_n$.
Take any $\alpha \in S_n$. Then either $\alpha \in A_n$ or not. Observe that, by the definition of $\phi$ and $\alpha^{\prime}$ in either case the image is an even permutation in $S_{n+2}.$ Can you see that?

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