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

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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 $G$).

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

Examples

Examples of index two

Group partSubgroup partQuotient part
A3 in S3Symmetric group:S3Cyclic group:Z3Cyclic group:Z2
Cyclic maximal subgroup of dihedral group:D16Dihedral group:D16Cyclic group:Z8Cyclic group:Z2
Cyclic maximal subgroup of dihedral group:D8Dihedral group:D8Cyclic group:Z4Cyclic group:Z2
Cyclic maximal subgroup of semidihedral group:SD16Semidihedral group:SD16Cyclic group:Z8Cyclic group:Z2
D8 in D16Dihedral group:D16Dihedral group:D8Cyclic group:Z2
D8 in SD16Semidihedral group:SD16Dihedral group:D8Cyclic group:Z2
Direct product of Z4 and Z2 in M16M16Direct product of Z4 and Z2Cyclic group:Z2
First omega subgroup of direct product of Z4 and Z2Direct product of Z4 and Z2Klein four-groupCyclic group:Z2
Klein four-subgroups of dihedral group:D8Dihedral group:D8Klein four-groupCyclic group:Z2
Non-normal subgroups of dihedral group:D8Dihedral group:D8Cyclic group:Z2
Q8 in SD16Semidihedral group:SD16Quaternion groupCyclic group:Z2
Q8 in central product of D8 and Z4Central product of D8 and Z4Quaternion groupCyclic 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 V4Klein four-groupCyclic group:Z2Cyclic group:Z2
Z4 in direct product of Z4 and Z2Direct product of Z4 and Z2Cyclic group:Z4Cyclic group:Z2

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 $H$ 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. 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 $G$ 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 $p$.