Subgroup of index equal to least prime divisor of group order is normal

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Statement

Let be a finite group and be the least prime divisor of the order of . Then, if is a subgroup of such that the index equals , then (i.e., is normal in ).

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Subgroup of index equal to least prime divisor of group order (?)) must also satisfy the second subgroup property (i.e., Normal subgroup (?)). In other words, every Subgroup of index equal to least prime divisor of group order of finite group (?) is a Normal subgroup of finite group (?).
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This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
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This result relates to the least prime divisor of the order of a group. View more such results

Related facts

Examples

Examples of index two

  Group part Subgroup part Quotient part
A3 in S3 Symmetric group:S3 Cyclic group:Z3 Cyclic group:Z2
Cyclic maximal subgroup of dihedral group:D16 Dihedral group:D16 Cyclic group:Z8 Cyclic group:Z2
Cyclic maximal subgroup of dihedral group:D8 Dihedral group:D8 Cyclic group:Z4 Cyclic group:Z2
Cyclic maximal subgroup of semidihedral group:SD16 Semidihedral group:SD16 Cyclic group:Z8 Cyclic group:Z2
D8 in D16 Dihedral group:D16 Dihedral group:D8 Cyclic group:Z2
D8 in SD16 Semidihedral group:SD16 Dihedral group:D8 Cyclic group:Z2
Direct product of Z4 and Z2 in M16 M16 Direct product of Z4 and Z2 Cyclic group:Z2
First omega subgroup of direct product of Z4 and Z2 Direct product of Z4 and Z2 Klein four-group Cyclic group:Z2
Klein four-subgroups of dihedral group:D8 Dihedral group:D8 Klein four-group Cyclic group:Z2
Non-normal subgroups of dihedral group:D8 Dihedral group:D8 Cyclic group:Z2
Q8 in SD16 Semidihedral group:SD16 Quaternion group Cyclic group:Z2
Q8 in central product of D8 and Z4 Central product of D8 and Z4 Quaternion group Cyclic group:Z2
SL(2,3) in GL(2,3) General linear group:GL(2,3) Special linear group:SL(2,3) Cyclic group:Z2
Z2 in V4 Klein four-group Cyclic group:Z2 Cyclic group:Z2
Z4 in direct product of Z4 and Z2 Direct product of Z4 and Z2 Cyclic group:Z4 Cyclic group:Z2

Examples of other prime index

These include:

Caution

The statement is that if we have a subgroup whose index is the least prime divisor of the order of the group, that subgroup is normal. The statement does not say that among the subgroups of prime index, the one of least prime index is normal. For instance, in the alternating group on five letters, there is no subgroup of index two (the least prime divisor). There is also no subgroup of index three. There are subgroups of index five, namely A4 in A5, and these are not normal.

Facts used

  1. Group acts on left coset space of subgroup by left multiplication: If is a subgroup of , then acts by left multiplication on the left coset space , yielding a homomorphism . The kernel of this homomorphism is the normal core of : the unique largest normal subgroup of contained in .
  2. Lagrange's theorem: The order of a subgroup divides the order of the group.
  3. Order of quotient group divides order of group

Proof

Proof using action on coset space

Given: A group and a subgroup such that , where is the least prime divisor of the order of .

To prove: is normal in .

Proof:

  1. (Facts used: fact (1)): Consider the action of on the left coset space of , by left multiplication (Fact (1)). This gives a homomorphism where is the symmetric group on , which has size . The kernel of this homomorphism is a normal subgroup of contained inside (in fact, it is the normal core of ).
  2. (Facts used: fact (2)):The image is a subgroup of , and hence, by fact (2), its order divides the order of . Thus, the order of divides .
  3. (Facts used: fact (3)): The image is isomorphic to the quotient group , and thus, by fact (3), its order divides the order of . Thus, the order of divides the order of .
  4. (Give data used: is the least prime divisor of the order of ): Since is the least prime divisor of the order of , we conclude that . Combining this with steps (2) and (3), we see that the order of divides . Since , we obtain that .
  5. We thus have that , with and . This forces , yielding that is a normal subgroup of .

Proof using action on the set of conjugates

Now, since is a maximal subgroup in , is either normal or self-normalizing. Assume by contradiction that is not normal. Then it is self-normalizing. The same is true for .

Consider the set of all conjugates of in . Then, acts on by conjugation. Restricting to , acts on by conjugation.

Thus, every element of cannot normalize , and hence the action of on has no fixed points other than itself.

We further know that the total cardinality of is , and that there is exactly one fixed point. Thus, there is a nontrivial orbit under whose size is strictly less than . But from the fact that the size of any orbit must divide the size of the group, we have a nontrivial divisor of the order of that is strictly smaller than , contradicting the least prime divisor assumption on .