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I'm new to Group theory and I'm just checking on my understanding. One example of 5-cycle is $(1\ 2\ 3\ 4\ 5)$. Hence, a subgroup generated by this 5-cycle consist of $\{(1\ 2\ 3\ 4\ 5), (1\ 3\ 5\ 2\ 4), (1\ 4\ 2\ 5\ 3), (1\ 5\ 4\ 3\ 2), e \}$, where $e$ is the identity element.

But what happens if it is generated by 2 5-cycles? E.g. $<(1\ 2\ 3\ 4\ 5),(1\ 4\ 3\ 5\ 2)>$. I start to get different cycles. One such element in this group is $(1\ 5\ 3)$.

Therefore, is there any generalizations I can obtain from the subgroups of $S_5$ generated by the 5-cycles?

How about if I were to extend the question to subgroups of $S_6$ generated by 6-cycles? Wouldn't it be more complicated to obtain some generalizations?

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The five-cycles are even permutations. So they all lie in $A_5$ and the group $G$ generated by two "independent" five-cycles is a subgroup of $A_5$. By Sylow's third theorem, as $G$ has at least two Sylow $5$-subgroups, it has at least six Sylow $5$-subgroups. There are only six Sylow $5$-subgroups in $A_5$ so $G$ must be the group generated by all the $5$-cycles. Therefore $G$ is normal in $A_5$.....

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  • $\begingroup$ So if I were to extend the question, in S_n, the n-cycles are even permutations if n is odd, and the n-cycles are odd permutations if n is even. Am I right? $\endgroup$ – Icycarus Oct 9 '18 at 21:11
  • $\begingroup$ @Icycarus That's right. As you notice, $n$ being prime is also very convenient... $\endgroup$ – Lord Shark the Unknown Oct 9 '18 at 21:13
  • $\begingroup$ Why is that so? $\endgroup$ – Icycarus Oct 9 '18 at 21:14
  • $\begingroup$ If I were to look at the 6-cycles, the 6-cycles are odd permutations. But if the subgroup is generated by 6-cycles, we cannot conclude that all the elements lie outside of A_6, is that right too? Since the product of odd permutations is an even permutation. $\endgroup$ – Icycarus Oct 9 '18 at 21:44

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