Congruence condition fails for number of normal subgroups of given prime power order

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Statement

It is possible to have a finite group G, and a prime power p^r dividing the order of G, such that the number of normal subgroups of G of order p^r is a nonzero number that is not congruent to 1 mod p.

Note that any example must have G as not being a finite p-group itself, because of the congruence condition on number of subgroups of given prime power order and the equivalence of definitions of universal congruence condition.

Related facts

Opposite facts

Proof

Further information: direct product of A4 and Z4, direct product of A4 and V4

Example of direct product of A4 and Z4

Let G be the group direct product of A4 and Z4. This is a group of order 48 (GAP ID: (48,31)) obtained as the external direct product of alternating group:A4 (order 12) and cyclic group:Z4 (order 4). Suppose we denote:

G = H_1 \times H_2

where H_1 is alternating group:A4 and H_2 is cyclic group:Z4.

Note that the order is of the form:

\! 48 = 2^4 \cdot 3

The group has exactly two normal subgroups of order 4:

  • The subgroup K \times \{ e \} where K is the subgroup in H_1 corresponding to V4 in A4, i.e., the subgroup \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}. This subgroup is isomorphic to a Klein four-group.
  • The subgroup \{ e \} \times H_2. This subgroup is isomorphic to cyclic group:Z4.

Thus, the number of subgroups of order 2^2 is 2, which is a nonzero number that is not congruent to 1 mod 2.

Example of direct product of A4 and V4

Let G be the group direct product of A4 and Z4. This is a group of order 48 (GAP ID: (48,31)) obtained as the external direct product of alternating group:A4 (order 12) and Klein four-group (order 4). Suppose we denote:

G = H_1 \times H_2

where H_1 is alternating group:A4 and H_2 is Klein four-group.

Note that the order is of the form:

\! 48 = 2^4 \cdot 3

The group has exactly two normal subgroups of order 4:

  • The subgroup K \times \{ e \} where K is the subgroup in H_1 corresponding to V4 in A4, i.e., the subgroup \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}. This subgroup is isomorphic to a Klein four-group.
  • The subgroup \{ e \} \times H_2. This subgroup is isomorphic to Klein four-group.

Thus, the number of subgroups of order 2^2 is 2, which is a nonzero number that is not congruent to 1 mod 2.