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I am wondering if there is any general relation between order of a group and order of its normal subgroups? In the problem that I am looking at, I know the prime factorization of the order of a group G and I know it has a normal subgroup N. I am able to relate that two using Lagrange theorem that |G/N| = |G|/|N|, but does that infer that the quotient group can take any order such that |N| divides |G|, i.e. the factors of |G|? Can someone please help me check if I am on the right track? Thanks!

Henno Brandsma
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1642920877
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  • If $G$ and $N$ is fixed, then there is only one value for $|G/N|$. I don't really understand what you're asking. – anon Mar 30 '21 at 01:04
  • Thank you for your reply! I am wondering if |N| is not fixed, can |G/N| just be any divisor of |G|? Or is there any restrictions on the order of a normal subgroup? – 1642920877 Mar 30 '21 at 01:12
  • It's unclear why you're restricting to normal subgroups - you don't need to do that. Of course $|G/N|$ will be a divisor of $|G|$, but in general there will exist divisors $d$ for which $d\ne |G/H|$ for any subgroup $H$ (that is, not all divisors are realizable). Really, you might as well ask if $|H|$ can be any divisor of $n$, since that is an equivalent question. See this question for more information and keywords to search. – anon Mar 30 '21 at 01:16

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Without knowing more information about the group it is impossible.

Notice that finite abelian groups have normal subgroups of every order that is a divisor of the order of the group ( because every subgroup is a normal subgroup in abelian groups). It can be a bit tricky to show this is true for arbitrary abelian groups, but it is easier to show for the cyclic group.

Asinomás
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Consider $A_4$, the fourth alternating group. It has no subgroup of index two.

On the other hand, it's easy to see that finite abelian groups are CLT groups. That's they satisfy the converse to Lagrange's theorem. (That's if follows easily from the structure theorem.)