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Following are orders of some proper subgroup of group. Which subgroup is necessarily normal?

$a)5. $ $ b)2.$ $ c)3. $ $d)4. $

Here order of group is not given so how to decide which is the answer? I know if index of a subgroup is 2 or p if p is smallest prime dividing the order of group then that subgroup is normal.But to calculate index, we need order of group but it is not given.I don't know how to proceed further.

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closed as off-topic by Alan Wang, Namaste, Claude Leibovici, Shailesh, user8795 May 28 '17 at 8:02

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  • $\begingroup$ Are you sure you didn't mean "following are the indexes...? Otherwise none is true. $\endgroup$ – DonAntonio May 27 '17 at 7:39
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None of the choices are necessarily normal.

Consider $A_5$, the alternating group of even permutations on $5$ elements.

Since $|A_5|=60$, $A_5$ has cyclic subgroups of orders $2,3,5$, but those subgroups can't be normal, since $A_5$ is a simple group.

Similarly, consider $A_8$, the alternating group of even permutations on $8$ elements. Then the product of two disjoint $4$-cycles of $S_8$ is an even permutation of order $4$, hence $A_8$ contains a cyclic subgroup of order $4$, which can't be normal since $A_8$ is a simple group.

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  • $\begingroup$ Why didn't you take any subgroup of order $\;4\;$ in $\;A_5\;$ itself again? Why $\;A_8\;$ all of a sudden for order $\;4\;$ ? $\endgroup$ – DonAntonio May 27 '17 at 7:40
  • $\begingroup$ There are no cyclic subgroups of order $4$ in $A_5$. Perhaps there is a subgroup of $A_5$ of order $4$ that is not cyclic -- I ddn't check, but it was easier to use $A_8$. $\endgroup$ – quasi May 27 '17 at 7:41
  • $\begingroup$ Why would we care about cyclic or non-cyclic? $\endgroup$ – DonAntonio May 27 '17 at 7:43
  • $\begingroup$ Because it's easier to compute the order without any work. $\endgroup$ – quasi May 27 '17 at 7:43
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    $\begingroup$ To downvote a correct answer is ridiculous... I shall upvote to compensate. +1 $\endgroup$ – DonAntonio May 27 '17 at 9:41

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