When proving that [tex]A_n[/tex] with [tex]n \geq 5[/tex] is simple, we require the following lemma:

If N is a normal subgroup of [tex]A_n[/tex] with [tex]n \geq 5[/tex] and N contains a 3-cycle, then [tex]N = A_n[/tex].

The proof is actually given for us in the lecture notes, however he utilizes a formula that I'm not sure how he derived:

Let [tex]f \in S_n[/tex] be a permutation and let [tex]x = (i_1 i_2 ... i_k)[/tex] be a k-cycle. Then:

[tex] fxf^{-1} = f(i_1 i_2 ... i_k)f^{-1} = ( f(i_1) f(i_2) ... f(i_k) ) [/tex]

where the right side is a cycle. I'm blind to how he gets from left to right.. and we were left with the ubiquitous "it is easy to check that..."

Any help to get started on seeing how this is true would be appreciated. Thanks,

Coto

If N is a normal subgroup of [tex]A_n[/tex] with [tex]n \geq 5[/tex] and N contains a 3-cycle, then [tex]N = A_n[/tex].

The proof is actually given for us in the lecture notes, however he utilizes a formula that I'm not sure how he derived:

Let [tex]f \in S_n[/tex] be a permutation and let [tex]x = (i_1 i_2 ... i_k)[/tex] be a k-cycle. Then:

[tex] fxf^{-1} = f(i_1 i_2 ... i_k)f^{-1} = ( f(i_1) f(i_2) ... f(i_k) ) [/tex]

where the right side is a cycle. I'm blind to how he gets from left to right.. and we were left with the ubiquitous "it is easy to check that..."

Any help to get started on seeing how this is true would be appreciated. Thanks,

Coto

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