# 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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## 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 .