# Order of normal subgroup [closed]

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.

## closed as off-topic by Alan Wang, Namaste, Claude Leibovici, Shailesh, user8795May 28 '17 at 8:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alan Wang, Namaste, Claude Leibovici, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

• Are you sure you didn't mean "following are the indexes...? Otherwise none is true. – DonAntonio May 27 '17 at 7:39

## 1 Answer

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.

• Why didn't you take any subgroup of order $\;4\;$ in $\;A_5\;$ itself again? Why $\;A_8\;$ all of a sudden for order $\;4\;$ ? – DonAntonio May 27 '17 at 7:40
• 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 $A_8$. – quasi May 27 '17 at 7:41
• Why would we care about cyclic or non-cyclic? – DonAntonio May 27 '17 at 7:43
• Because it's easier to compute the order without any work. – quasi May 27 '17 at 7:43
• To downvote a correct answer is ridiculous... I shall upvote to compensate. +1 – DonAntonio May 27 '17 at 9:41