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My attempt : By Lagrange's theorem the order of $H$ should be 6.So $H$ should be generated by ($a$)6 cycle, ($b$) 3cycle and 2 cycle. I think these are the only cases of $H$.

Now, if $H=<(a_1...a_6)>$ , then $gHg^{-1} \in H$.As $H$ is a normal subgroup, we see that $g(a_1)..g(a_6) \in H$.Now, there are $120$ elements of 6 cycle.As the elements of cycle $6$ are conjugates of each other, we can pick up an element $c$ such that $g(a_1..a_6)g^{-1} =c$ where$ c \ne\in H$ .We are done.

Now,$<(a_1 a_2)(a_3 a_4 a_5)>=H$ so how do we progress with this case.

Guria Sona
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  • No the index that I want should be 5! And not 2 – Guria Sona Sep 14 '20 at 07:16
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    Clearly if $S_n$ has only $A_n$ as a proper non-trivial normal subgroup, which is always of index $2$, then $S_6$ has no quotient isomorphic to $S_5$. In fact there are only 3 quotients of $S_n$: these are the trivial group $E$, $S_n$ itself and $\mathbb{Z}_2$. – freakish Sep 14 '20 at 07:18
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    The link says that there is no normal subgroup with index $\not\in{1, 2,6!}$. – Vercassivelaunos Sep 14 '20 at 07:19
  • Since a normal subgroup of $S_6$ consists of complete conjugacy classes, you can count the conjugates of the two possible generators you have identified and show that there are more than five of each (add the identity too). $H$ would have to contain at least these elements to be normal. – Mark Bennet Sep 14 '20 at 07:27
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    See also your former question. – Dietrich Burde Sep 14 '20 at 09:15
  • My answer to your former question generalises to all $S_n$ with $n$ not a power of two. So it also works for $n=6$. – user1729 Sep 14 '20 at 14:21

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If $H$ contains an element $h$ which is the product of a $2$-cycle and a $3$-cycle, it contains the $3$-cycle $h^2$.

And as $H$ is supposed to be normal, it contains all the $3$-cycles (all the $3$-cycles are conjugate elements). As the $3$-cycles generate $A_6$, $H$ contains $A_6$. A contradiction.

mathcounterexamples.net
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