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

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.

Opposite facts

 * Congruence condition on number of subgroups of given prime power order

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.