# Collection of groups satisfying a universal congruence condition

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## Definition

Suppose $\mathcal{S}$ is a finite collection of finite $p$-groups, groups of prime power order for the prime $p$. We say that $\mathcal{S}$ satisfies a universal congruence condition if the following equivalent conditions are satisfied by $\mathcal{S}$:

1. For any finite $p$-group $P$ that contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of subgroups of $P$ isomorphic to elements of $\mathcal{S}$ is congruent to $1$ modulo $p$.
2. For any finite $p$-group $P$ that contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of normal subgroups of $P$ isomorphic to elements of $\mathcal{S}$ is congruent to $1$ modulo $p$.
3. For any finite $p$-group $Q$ and any normal subgroup $P$ of $Q$ such that $P$ contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of normal subgroups of $Q$ isomorphic to elements of $\mathcal{S}$ and contained in $P$ is congruent to $1$ modulo $p$.
4. For any finite $p$-group $P$ that contains a subgroup isomorphic to an element of $\mathcal{S}$, the number of p-core-automorphism-invariant subgroups of $P$ isomorphic to elements of $\mathcal{S}$ is congruent to $1$ modulo $P$.
5. For any finite group $G$ containing a subgroup isomorphic to an element of $\mathcal{S}$, the number of subgroups of $G$ isomorphic to an element of $\mathcal{S}$ is congruent to $1$ modulo $p$.

### Equivalence of definitions

Further information: equivalence of definitions of universal congruence condition

## Examples/facts

### Satisfaction

Collection Conditions on prime $p$ Conditions on $k$ Proof
All groups of order $p^k$ all $p$ all $k$ congruence condition on number of subgroups of given prime power order
Elementary abelian group of order $p^k$ odd prime $0 \le k \le 5$ Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime
Abelian groups of order $p^k$ odd prime $0 \le k \le 5$ Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime
Abelian groups of order $p^k$, exponent dividing $p^d$ odd prime $0 \le d \le k \le 5$ Congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime
Abelian groups of order $p^3$ all $p$ $k = 3$ congruence condition on number of abelian subgroups of prime-cube order
Abelian groups of order $p^4$ all $p$ $k = 4$ congruence condition on number of abelian subgroups of prime-fourth order
Abelian groups of order $8$, exponent dividing $4$ $p = 2$ $k = 3$ congruence condition on number of abelian subgroups of order eight and exponent dividing four
Abelian groups of order $16$, exponent dividing $8$ $p = 2$ $k = 4$ congruence condition on number of abelian subgroups of order sixteen and exponent dividing eight
Non-cyclic groups of order $p^3$ odd $p$ $k = 3$ congruence condition on number of non-cyclic subgroups of prime-cube order for odd prime

### Dissatisfaction

Collection Conditions on prime $p$ Conditions on $k$ Proof
Klein four-group $p = 2$ $k = 2$ elementary abelian-to-normal replacement fails for Klein four-group
Elementary abelian group of order $2^k$ $p = 2$ $k \ge 2$ Follows from above
Elementary abelian group of order $p^k$ all $p$ $k \ge 6$
Abelian groups of order $p^6$ all $p$ $k = 6$ Congruence condition fails for abelian subgroups of prime-sixth order
Abelian groups of order $p^k$ all $p$ $k \ge 6$ Follows from above.
Elementary abelian group of order $p^6$ all $p$ $k = 6$ Congruence condition fails for elementary abelian subgroups of prime-sixth order (example same as for abelian subgroups)
Elementary abelian group of order $p^k$ all $p$ $k \ge 6$ Follows from above.
Groups of order $p^p$, exponent $p$ all $p$ $k = p$ Congruence condition fails for subgroups of order p^p and exponent p

### Threshold values

This lists threshold values of $k$: the largest value of $k$ for which the collection of $p$-groups of order $p^k$ satisfying the stated condition satisfies a universal congruence condition. The nature of all these is such that the universal congruence condition is satisfied for all smaller $k$ but for no larger $k$. We use between $a$ and $b$ to mean that the value is at least $a$ and at most $b$.

Collection of groups $p = 2$ $p = 3$ $p = 5$ $p = 7$ $p \ge 11$
Abelian groups of order $p^k$ between 4 and 5 5 5 5 5
Abelian groups of order $p^k$, exponent dividing $p^d$, $2 \le d \le k$ between 3 and 5 5 5 5 5
Elementary abelian group of order $p^k$ 1 5 5 5 5
Groups of exponent $p$, order $p^k$ 1 2 between 2 and 4 between 2 and 6 between 2 and $p - 1$