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

Statement


Let $$G$$ be a finite group and $$p$$ be the least prime divisor of the order of $$G$$. Then, if $$H$$ is a subgroup of $$G$$ such that the index $$[G:H]$$ equals $$p$$, then $$H \underline{\triangleleft} G$$ (i.e., $$H$$ is normal in $$G$$).

Related facts

 * 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 $$n$$ in a group contains a normal subgroup of finite index (with the index a multiple of $$n$$ and a divisor of $$n!$$). 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 other prime index
These include:


 * 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.

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) uses::Group acts on left coset space of subgroup by left multiplication: If $$H$$ is a subgroup of $$G$$, then $$G$$ acts by left multiplication on the left coset space $$G/H$$, yielding a homomorphism $$G \to \operatorname{Sym}(G/H)$$. The kernel of this homomorphism is the normal core of $$H$$: the unique largest normal subgroup of $$G$$ contained in $$H$$.
 * 2) uses::Lagrange's theorem: The order of a subgroup divides the order of the group.
 * 3) uses::Order of quotient group divides order of group

Proof using action on coset space
Given: A group $$G$$ and a subgroup $$H$$ such that $$[G:H] = p$$, where $$p$$ is the least prime divisor of the order of $$G$$.

To prove: $$H$$ is normal in $$G$$.

Proof:
 * 1) (Facts used: fact (1)): Consider the action of $$G$$ on the left coset space of $$H$$, by left multiplication (Fact (1)). This gives a homomorphism $$\rho: G \to S_p$$ where $$S_p$$ is the symmetric group on $$G/H$$, which has size $$p$$. The kernel of this homomorphism is a normal subgroup $$N$$ of $$G$$ contained inside $$H$$ (in fact, it is the normal core of $$H$$).
 * 2) (Facts used: fact (2)):The image $$\rho(G)$$ is a subgroup of $$S_p$$, and hence, by fact (2), its order divides the order of $$S_p$$. Thus, the order of $$\rho(G)$$ divides $$p!$$.
 * 3) (Facts used: fact (3)): The image $$\rho(G)$$ is isomorphic to the quotient group $$G/N$$, and thus, by fact (3), its order divides the order of $$G$$. Thus, the order of $$\rho(G)$$ divides the order of $$G$$.
 * 4) (Give data used: $$p$$ is the least prime divisor of the order of $$G$$): Since $$p$$ is the least prime divisor of the order of $$G$$, we conclude that $$\operatorname{gcd}(p!, |G|) = p$$. Combining this with steps (2) and (3), we see that the order of $$\rho(G)$$ divides $$p$$. Since $$\rho(G) \cong G/N$$, we obtain that $$[G:N] | p$$.
 * 5) We thus have that $$N \le H \le G$$, with $$[G:H] = p$$ and $$[G:N] | p$$. This forces $$N = H$$, yielding that $$H$$ is a normal subgroup of $$G$$.

Proof using action on the set of conjugates
Now, since $$H$$ is a maximal subgroup in $$G$$, $$H$$ is either normal or self-normalizing. Assume by contradiction that $$H$$ is not normal. Then it is self-normalizing. The same is true for $$H^g$$.

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

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

We further know that the total cardinality of $$S$$ is $$[G:N_G(H)] = p$$, and that there is exactly one fixed point. Thus, there is a nontrivial orbit under $$H$$ whose size is strictly less than $$p$$. 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 $$H$$ that is strictly smaller than $$p$$, contradicting the least prime divisor assumption on $$p$$.