A Useful Property of the Finite Nonabelian Groups

Definition 1 Let G be a finite nonabelian group with center Z and let the r maximally abelian subgroups be

for all i, j, i = j and G is called Z-dependent if there exists i, j i = j and H i ∩ H j = D ⊃ Z and D = Z.

Theorem 2 Let G be a finite nonabelian group with center Z and let H = {H 1 , H 2 , • • • , H r } be the set of the r maximally abelian subgroups of G. Then the following 6 claims hold:

Let G be Z-dependent and suppose that

is the subset of all maximally abelian subgroups of G which contain subgroup D ⊃ Z, D = Z and let

Moreover, the maximally abelian subgroups of < H D > are those of G containing D and are thus in

∈ HZ and HZ is a subgroup of G. But HZ is also abelian for (vi) Otherwise G is the union of two proper subgroups, which is impossible.

We need a definition.

Theorem 4 Let G be a finite group with eigenheimer H and let x ∈ G. Then the conjugate subgroup xHx -1 is also an eigenheimer of G.

) and the equality follows.

We prove a few grouptheoretic theorems:

Theorem 5 Let G be a finite group and let H be a proper subgroup of G Then

The Main Theorem for Finite Nonabelian Groups 6 Let G be a finite nonabelian group with center Z. Then one of the maximally abelian subgroups of G is not an eigenheimer of G.

Proof: We use induction on the number |G| and we assume that G is a minimal counterexample. There are two cases to be considered: (i) Firstly that G is Z-independent. Let H and K be maximally abelian subgroups of G and let H = K. Then H ∩K = Z. We assume that all maximally abelian subgroups are eigenheimers, otherwise we are done. Let |G| = g, |H| = h, |K| = k and |Z| = z. Suppose (ia) that all maximally abelian subgroups are conjugate with H. Then gz = g h (hz) or g = h. A contradiction. Suppose (ib) H and K are not conjugate. Then (gz)

From these two contradictions it follows that at least one of the maximally abelian subgroups, let us say H, is not an eigenheimer: N G (H) = H.

(ii) Secondly that G is Z-dependent. Referring to theorem 2 (v) and knowing that the subgroup < H D > = G we use induction on | < H D > |. As the maximally abelian subgroups of < H D > are maximally abelian subgroups of G, then by induction one of the H ′ i , i = 1, 2, • • • , s is not an eigenheimer in < H D > and thus also not in G. This completes the proof.

Theorem 7 Let G be a finite nonabelian group such that the proper subgroups of G are all abelian. Then one of the maximal subgroups of G is an abelian normal subgroup of G.

Proof: One of the maximal subgroups, which is abelian, is not an eigenheimer and thus a normal subgroup of G.

Let G be a finite group and for every abelian subgroup H of G we have N G (H) = C G (H). Then G is abelian.

Proof: Suppose that G is nonabelian. Let H i be the maximally abelian subgroups of G. Then one of the maximally abelian subgroups, let us say H, is not an eigenheimer by theorem 6. Thus

then the group generated by H and c is abelian. A contradiction. Thus G is abelian.

(This was also the original title of version v1, so I’m happy again.)

We will prove the well-known fact that a finite division ring is commutative (Wedderburn, 1905) along the following lines. We start with a minimal counterexample L, with center Z, viz. a finite division ring which is not commutative but all its maximal division subrings are commutative and are thus subfields. We say that L is Z-dependent or Z-independent when we mean actually that L × is Z × -dependent or Z × -independent. There are two possibilities to be considered. The first case (I) is where L is Z-dependent and the second case (II) is where L is Z-independent, see Definition 1.

be the set of the r maximal subfields of the finite division ring L. All the maximal subfields contain the center Z of L and let

be the set of maximal subfields containing also D. Then F ′ is not only a multiplicative group (if we drop the zero) (see Theorem 2 (v)) but also an additive subgroup (if we allow the zero) and thus a proper division ring which is by induction commutative as we now will show the additive case. Let

be the set of the r maximal subfields of the finite division ring L. All the maximal subfields contain the center Z of L. We assume now that F i ∩ F j = Z for all i and j, i = j and 1 ≤ i, j ≤ r. Thus L is Z-independent. According to main theorem 6 one of the maximally abelian subgroups F × i , i = 1, • • • , r is not an eigenheimer. Let us say that F × 1 is not an eigenheimer so that N := N L × (F × 1 ) = F × 1 . We assume that |F 1 | = p m and |Z| = p z so that z|m. Let zs|m be such that s is a prime number then F 1 contains a subfield E such that |E| = p zs . But then also N = N L × (E × ), as we shall now prove. Let us recall that a finite field minus the zero is a cyclic group, a subgroup of a finite cyclic group is also cyclic and that two subgroups of a finite cyclic group with the same order are equal and moreover that the quotient group of a finite cycli