Subgroup of index equal to least prime divisor of group order is normal
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
- Subgroup of index two is normal: This statement is true for any group, whether finite or infinite.
- Nilpotent implies every maximal subgroup is normal. Further, any maximal subgroup in finite nilpotent group is a normal subgroup of prime index, i.e., the quotient is a group of prime order. In particular, since prime power order implies nilpotent, it is true that in a group of prime power order, the maximal subgroups are precisely the same as the maximal normal subgroups, and these are precisely the same as the subgroups of prime index.
- Homomorphism between groups of coprime order is trivial: The proof uses the same idea: a combination of Lagrange's theorem for subgroups and for quotient groups.
- Poincare's theorem: This states that every subgroup of finite index in a group contains a normal subgroup of finite index (with the index a multiple of and a divisor of ). The normal subgroup of finite index can be chosen as the normal core.
- Normal of order equal to least prime divisor of group order implies central
Examples of index two
Examples of other prime index
- All examples of maximal subgroups of groups of prime power order.
- The normal subgroup isomorphic to cyclic group:Z7 in the semidirect product of Z7 and Z3.
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.
- 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 .
- Lagrange's theorem: The order of a subgroup divides the order of the group.
- Order of quotient group divides order of group
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 .
- (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 ).
- (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 .
- (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 .
- (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 .
- 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 .