On Group Violations of Inequalities in five Subgroups

We consider ten linear rank inequalities, which always hold for ranks of vector subspaces, and look at them as group inequalities. We prove that groups of order pq, for p,q two distinct primes, always satisfy these ten group inequalities. We give partial results for groups of order $p^2q$, and find that the symmetric group $S_4$ is the smallest group that yield violations, for two among the ten group inequalities.

Let G be a finite group, and G 1 , . . . , G n be subgroups of G. Then is a subgroup obtained by intersecting G i , i ∈ A. Let X be a random variable uniformly distributed over G: P (X = g) = 1 |G| , g ∈ G. Then the random variable X i = XG i whose support is the [G : G i ] cosets of G i in G satisfies P (X i = gG i ) = |Gi| |G| (see [1]) and The entropy of the random variable X A = (X i , i ∈ A), X i = XG i , is In what follows, we will only consider random variables X 1 , . . . , X n coming from a finite group G and n of its subgroups G 1 , . . . , G n . Hence, to any information (in)equality (that is, (in)equality written as a function of joint Shannon entropies) expressed in terms of random variables X 1 , . . . , X n will correspond a group (in)equality which is a function of the subgroups G 1 , . . . , G n , using H(X A ) = log |G| |G A | . We can drop the logarithm from an inequality by exponentiating both sides. We will denote this correspondence from linear inequalities into the multiplicative group inequalities by G.
• The mutual information I(X 1 ; X 2 ) = H(X 1 ) + H(X 2 ) − H(X 1 , X 2 ) becomes • The conditional mutual information 1 I(X 1 ; X 2 |X 3 ) = H(X 1 |X 3 ) − H(X 1 |X 2 , X 3 ) is similarly Linear inequalities satisfied by any joint distribution of random variables, such as the non-negativity of the conditional mutual information are called information inequalities. Similarly, linear inequalities satisfied by ranks of any vector subspace arrangement are called linear rank inequalities.
A vector space V over a finite field F, along with its subspaces V 1 , . . . , V n gives rise to random variables in the following fashion. Let X be a uniform random variable over the set of linear functions from V to F. Further let X i be the restriction X| Vi for each i = 1, . . . , n. Each X i is a uniform random variable with |F| dim(Vi) values, so H(X| Vi ) = dim(V i ) log |F|. This shows [5,Theorem 2] how information inequalities are a subset of linear rank inequalities. The converse is generally not true.
Consider the case of n = 4 random variables. The Ingleton inequality [7] I(X 1 ; X 2 ) ≤ I(X 1 ; X 2 |X 3 ) + I(X 1 ; X 2 |X 4 ) + I(X 3 ; X 4 ) always holds for ranks of subspaces, yet there are examples of random variables whose joint entropies violate the Ingleton inequality [11]. In the context of ranks of subspaces, H(X 1 ) is understood as the rank of the subspace X 1 , H(X 1 , X 2 ) as the rank of the span X 1 , X 2 , and I(X 1 ; X 2 ) as the rank of X 1 ∩ X 2 , thus the fact that the mutual information I(X 1 ; X 2 ) satisfies I(X 1 ; X 2 ) = H(X 1 ) + H(X 2 ) − H(X 1 , X 2 ) corresponds to the well-known formula that relates the rank of a span of two subspaces to that of their intersection. The idea of looking at violations of the Ingleton inequality coming from groups was proposed in [9]. By a group violation, we mean a group G, with n subgroups G 1 , . . . , G n , such that the orders of the intersections of subgroups involved do not satisfy a given inequality. It was shown in [9] that the symmetric group S 5 and four of its subgroups G 1 , . . . , G 4 violate the group version of the Ingleton inequality At the same time, it is also known [1] that abelian groups always satisfy the Ingleton inequality. The Ingleton inequality plays a special role for n = 4 random variables. Together with the Shannon inequalities (a special case of information inequalities which are derived from the non-negativity of the conditional mutual information), the Ingleton inequality generates all linear rank inequalities in n = 4 random variables.
From an information theoretic view point, it is fundamental to understand the region Γ * n formed by (2 n −1)-dimensional vectors of the form (H(X A ), A ⊂ {1, . . . , n}) where A runs through all non-empty subsets of {1, . . . , n}. Indeed, it has been shown that the capacity region of any arbitrary multi-source multi-sink acyclic network can be computed as a linear function of such vectors under linear constraints [6,15]. But the problem of describing Γ * n is notoriously difficult. The subset of Γ * n consisting of entropy vectors obtained from linear subspaces (i.e those satisfying linear rank inequalities) on the other hand is too restricted, and only captures a small part of the region of interest. To further our understanding of the rest of Γ * n , we seek to find random variables violating linear rank inequalities. Using information theoretic tools only, this is not an easy task. Group theory provides a systematic way to address this question by considering the larger subset of R 2 n −1 consisting of elements of the form ( |G| |G A | , A ⊂ {1, . . . , n}) coming from groups [1]. From a practical point of view, the networks associated with this region may outperform those which are subject to linear constraints.
We therefore have the following hierarchy of the entropy region: entropies coming from subspaces ⊆ entropies coming from groups ⊆ entropies coming from all random variables. In case n = 4, the Ingleton inequality (along with Shannon-type inequalities) completely characterises those entropy vectors coming from subspaces.
Consider now the case of n = 5 random variables. In [2], it was shown that 24 inequalities, together with the Shannon inequalities, and three Ingleton-type inequalities, generate all linear rank inequalities on five variables. In this paper, we are interested in looking at these inequalities from a group theoretic point of view. The motivation is the same as that of [9] for four random variables, namely: finding violators furthers our understanding of Γ * n beyond the subregion characterised by linear rank inequalities.
Among the inequalities of [2], the three Ingleton-type inequalities have already been studied in [10], where it was shown that a group does not violate the Ingleton inequality (1) if and only if it does not violate any of the three Ingleton-type inequalities in 5 random variables. We thus focus on the first 10 out of the 24 inequalities of [2] (since they share some commonalities, or more precisely, the same form of common information, see Lemma 1 below).
In Section 2, we prove some general properties about the group inequalities corresponding to the first 10 inequalities of [2], that we refer to as DFZ inequalities throughout the paper.
In Section 3, we prove that groups of order pq, where p and q are two primes, never violate any of the DFZ inequalities. By characterizing groups which never yield violations, we make the search for violators tractable by reducing it to smaller families of groups.
Partial results are given in Section 4, namely we show that groups G of order p 2 q and subgroups (G 1 , G 2 , G 3 , G 4 , G 5 ) do not violate the ten DFZ inequalities as long as |G 1 | = |G 2 | = p does not hold. The case |G 1 | = |G 2 | = p remains open, and we discuss why it is more difficult than the other ones.
Violators are discussed in Section 5, where we show that the smallest violator of two of the ten DFZ inequalities is the symmetric group S 4 . We also show that it is not possible to find one group which simultaneously violate all the 10 DFZ inequalities.
Basic group theory is used in this paper, the interested reader may refer to [4] for a first introduction to group theory, to [12] for a course in group theory, and to [3] for general references about group theory.

Ten group inequalities
Recall that for n = 5 random variables, it was shown in [2] that 24 inequalities, together with the Shannon inequalities, and three Ingleton-type inequalities, generate all linear rank inequalities on 5 variables. The first 10 inequalities are The following lemma reveals our motivation to look at the 10 DFZ inequalities.

Lemma 1 ([2]
). Suppose there exists a random variable Z, called common information, such that then each of the 10 DFZ inequalities can be deduced from Shannon inequalities involving the random variables X 1 , . . . , X 5 .
The other 14 inequalities are also deduced from Shannon inequalities through the concept of common information, but it takes different expressions.
Repeating the computations of the previous section, inequalities (2)-(11) have a corresponding group theoretic formulation, obtained via the correspondence G: yielding accordingly 10 groups inequalities, which we will freely refer to as group inequalities (2)- (11). The notion of common information is also expressible in the language of groups: Lemma 2. If G 1 G 2 is a subgroup of G, then the group inequalities (2)-(11) hold.
Proof. We translate the above results on common information: similarly H(Z|X 2 ) = 0 ⇐⇒ G 2 < G Z and which shows the existence of common information and concludes the proof.
Corollary 1. The above 10 DFZ inequalities hold in the following cases: 3. G is abelian.
We will use the following lemma and corollary to prove many inequalities.
Lemma 3. Let G be a finite group with n subgroups G 1 , . . . , G n . For any choice of subsets Proof. We use the fact that the subset G A1∪A2 G A1∪A3 ≤ G A1 has cardinality . Moreover when one of G A1∪A2 , G A1∪A3 is normal, then the subset G A1∪A2 G A1∪A3 is actually a subgroup of G A1 , hence its size divides |G A1 |.
Relating the lemma above to information quantities i a;b|c = |G abc ||Gc| |Gac||G bc | and i a;b|c,d = |G abcd ||G cd | |G acd ||G bcd | gives the following immediate corollary. Corollary 2. Consider subgroups G a , G b , G c , G d ≤ G with corresponding random variables X a , X b , X c , X d . The following bounds hold: 1 ≤ i a;b|c and 1 ≤ i a;b|c,d .
Moreover, if either G ac G c or G bc G c then i a;b|c and i a;b|c,d are both integers.
Proof. Recall that i a;b|c = |G abc ||Gc| |Gac||G bc | and i a;b|c,d = |G abcd ||G cd | |G acd ||G bcd | . We apply the lemma above for subsets A 1 = {c}, A 2 = {a}, A 3 = {b} to show the first inequality. We apply the lemma above for subsets A 1 = {cd}, A 2 = {a}, A 3 = {b} to show the second inequality. The last statement is similarly a direct consequence of the lemma.
Next we show how uniqueness of a Sylow q-subgroup imposes constraints on quantities i a;b|c . We will use the valuation v q at a prime q. For integers u, v, let q eu and q ev be the powers of q that appear in the respective prime factorization of u and v.
Lemma 4. Let G be a group whose Sylow q-subgroup is normal and abelian. For . To see the last inequality note first that A q and B q are both contained in G q , since G q is normal (or equivalently unique). Moreover, since G q is by assumption abelian, A q B q in fact forms a subgroup, not just a subset of G q . Hence we know that |A q B q | divides |G q |.
As a corollary we get the following proposition. Proposition 1. Let G be a group whose Sylow q-subgroup is normal and abelian. Let G a , G b , G c be arbitrary subgroups of G. Then v q (i a;b|c ) ≥ 0.

Proof. Apply previous Lemma with
Remark 1. The fact that there is only one (normal) Sylow q-subgroup restricts powers of q in expression i a;b|c to positive ones. This eventually allows us to prove our results for groups of order pq and p 2 q in the following two sections.

Groups of order pq
In this section we prove that information inequalities (2)-(11) hold for groups of order pq .
First we make a few observations about these groups. The property described in the lemma below was used to prove inequalities (2)-(3) in [10].

Lemma 5.
A group G has the property that all its distinct proper subgroups have trivial intersection if and only if |G| = pq, where p, q are either 1 or prime, not necessarily distinct.

Proof. ⇐
Suppose |G| = pq and consider two proper distinct non-trivial subgroups A, B of G. If |A| = p and |B| = q for p = q, or vice-versa, they necessarily intersect trivially. If |A| = |B| = p (or q if p = q), then any non-trivial element of A ∩ B generates all of A and all of B, as they are of prime order -a contradiction since A, B are distinct.
⇒ Conversely, assume the property that all proper subgroups of G have trivial intersection. We will show that the order n of G is pq.
Case 1: G is equal to its Sylow p-subgroup. Then |G| = p t and it is a standard exercise using central series of G to show that G contains a subgroup p k for each 1 ≤ k ≤ t. It follows that if t > 2, then G contains a (proper) subgroup of order p 2 , which in turn has a subgroup of order p. These two subgroups have non-trivial intersection, contradicting our assumption on G. We conclude t ≤ 2.
Case 2: All Sylow subgroups are proper.
• We show that all Sylow subgroups are cyclic and hence n is a product of distinct primes. Consider a Sylow p-subgroup S p of order p t . It suffices to show that t = 1. Suppose, to the contrary, that t ≥ 2. Then by Cauchy theorem, S p contains a cyclic subgroup C p of order p, so S p and C p are two proper subgroups which intersect non-trivially -a contradiction. • Next we show that the number of divisors of n is at most 2. Suppose to the contrary, that the order n of G is divisible by at least three primes p, q, r.
Then pq is a Hall divisor of n -that is pq | n, gcd(pq, n pq ) = 1. Hence there exists a Hall-pq subgroup -a (proper) subgroup of order pq. But it will then intersect non-trivially with a Sylow p-subgroup -a contradiction.
Next we show that (2)-(11) hold for groups of order pq. The case |G| = p 2 is covered by abelian groups, which are known not to violate DFZ inequalities by Corollary 1. In what follows we hence assume that |G| = pq, with primes p < q.
Normality (equivalently, uniqueness) of Sylow q-subgroup plays an important role in establishing inequalities over the next two sections. Lemma 6. If |G| = pq, p < q, then G contains exactly one Sylow q-subgroup. Equivalently if a subgroup H ≤ G has order q, then it is necessarily normal.
Proof. This is an elementary exercise in group theory. Sylow's theorems tell that the number n q of Sylow q-subgroup divides pq and n q = 1 + kq for some k a positive integer, and also n q |G| = pq. Since p < q, the only choice left is n q = 1. The second claim follows from the fact that Sylow subgroups are conjugate of each other.
Corollary 3. Let |G| = pq, with primes p < q. If G 1 , G 2 ≤ G are subgroups of G, such that the set G 1 G 2 = {g 1 g 2 : g 1 ∈ G 1 , g 2 ∈ G 2 } does not form a subgroup, then G 1 , G 2 are distinct subgroups each of order p.
Proof. If the subgroups are not distinct, we have G 1 G 2 = G 1 is a subgroup. If one of G 1 , G 2 is normal, then the set G 1 G 2 forms a subgroup. But a subgroup of order other than p is necessarily normal and the result follows. Now we set out to prove inequalities (2)-(11) for groups of order pq. All these inequalities comprise mutual information terms, of the general form for a, b, c some indices or subsets in {1, 2, 3, 4, 5}. In group form, we have The next proposition shows how normality of the Sylow q-subgroup (guaranteed by Lemma 6) restricts information quantities i a;b|c . Proposition 2. Let G be a group of order pq, for primes p < q, so the Sylow q-subgroup is unique. Each quantity i a;b|c , i a;b|G , i a;b|c,d is ≥ 1 and takes values among {1, p, q p , q, pq}.
Proof. Let us make the argument in the case of the quantity i a;b|c = |G abc ||Gc| |Gac||G bc | , which is the same as |Gc| |GacG bc | . Since G ac G bc is a subset of G c , |G abc ||Gc| |Gac||G bc | ≥ 1. Since the Sylow q-subgroup is unique (equivalently, normal), by Proposition 1, v q (i a;b|c ) ≥ 0, so the powers of q occurring in i a;b|c can be only 0, 1. Now since G ac ≤ G c , G bc ≤ G c , the powers of p are restricted to −1, 0, 1, and The case 1 p cannot happen since i ≥ 1. The next two quantities follow as special cases. In case of the second quantity i a;b|G = |G ab ||G| |Ga||G b | , we use c = {}, i.e. G c = G. In case of the third quantity i a;b|c,d = |G abcd ||G cd | |G acd ||G bcd | , we just substitute G c with G cd and the same argument holds.
By abuse of notation write q | i a;b|c when v q (i a;b|c ) > 0, i.e., the quantity i a;b|c takes value among {q/p, q, pq} (even if some of these are not integers). Conversely we will write q i a;b|c , to indicate i a;b|c takes value among {1, p}.
Finally we make one small observation, which we will use a lot, thus it deserves to be stated as the following lemma.
Lemma 7. Let i denote one of the information quantities i a;b|c , i a;b|G or i a;b|c,d .
Suppose q i. Then we have Proof. Consider the first case i = |G abc ||Gc| |Gac||G bc | . By assumption q |G abc ||Gc| |Gac||G bc | . Rewrite |G abc ||Gc| |Gac||G bc | as |Gc| |GacG bc | . Depending on whether G ac G bc forms a subgroup, this may or may not be an integer. However, if q | |G c |, yet q |G abc ||Gc| |Gac||G bc | , then necessarily Taking the contrapositive gives the other implication q |G a ||G b | =⇒ q |G c |. The cases i a;b|G or i a;b|c,d follow, as discussed in the previous lemma, as special cases of the first case.
We are now ready to prove the main proposition of this section.
Proposition 3. Let G be a group of order pq, for primes p < q, so the Sylow q-subgroup is unique.
Proof. We start by proving in detail inequality (4), which in its unsimplified group form is where we have labeled the last three factors as I 1 = i 1;2|4 , I 2 = i 2;5|3 and I 3 = i 1;4|3,5 . We first prove the following claim about this inequality. Identical reasoning applies to other inequalities and we will simply refer to this claim when needing to invoke it.
Recall that by abuse of notation write q | RHS to mean that one of the factors on the RHS takes value among q/p, q, pq (even if some of these are not integers).

Claim 1. It suffices to show that q | RHS.
It has been shown in Lemma 2 that when G 1 G 2 is a subgroup of G such inequalities hold. Hence we only need to consider the case when the set G 1 G 2 is not a subgroup, which by Corollary 3 occurs when G 1 , G 2 are distinct, each of order p, so we have |G12||G| |G1||G2| = q p . The LHS is ( |G12||G| |G1||G2| ) = q p , while each of the factors on the right hand side takes values among {1, p, q/p, q, pq}. Clearly, if q divides one of the factors on the RHS, the inequality holds since each of the (other) factors is ≥ 1. This proves the claim.
Claim 2. We show that q | RHS.
Hence we showed that q | RHS, so by claim 1 the inequality is satisfied. We proceed in the same fashion for other inequalities. Each of these inequalities has the form where in each case I 1 , I 2 and I 3 will denote some information quantities on subgroups G 1 , G 2 , G 3 , G 4 , G 5 As before it suffices to show that one of the factors on the RHS takes value among { q p , q, pq}. Inequality (2).
with the last three factors now labeled as I 1 , I 2 , I 3 .
with the last four factors now labeled as I 1 , I 2 , I 3 , Inequality (9). This and subsequent inequalities have a different form: with the factors now labeled as I 1 , I 2 , I 3 , I 4 , I 5 .
The following claim is a generalization of claim 1 to inequalities of this type.
Claim 3. In order to prove the inequality, it suffices to show that the power of q dividing I 1 · I 2 · I 3 · I 4 · I 5 is at least 2.
The LHS is ( |G12||G| |G1||G2| ) = ( q p ) 2 , while each of the factors on the right hand side takes values among {1, p, q/p, q, pq}. Clearly if at least two of them are among q/p, q, pq, the inequality holds for each I 1 , I 2 , I 3 , I 4 , I 5 ≥ 1 by Proposition 2. This proves the claim.
Next we show that the condition of the claim is indeed satisfied.
Claim 4. The power of q dividing I 1 · I 2 · I 3 · I 4 · I 5 is at least 2.
So suppose q I 1 , q I 2 , q | I 4 . We must demonstrate that q divides at least one other factor.
with the factors now labeled as I 1 , I 2 , I 3 , I 4 , I 5 .
We conclude q I 1 =⇒ q | I 5 or (q | I 4 and q | I 3 ). Suppose q I 2 , then q | |G 4 ||G 5 | =⇒ q | I 3 · I 4 . q I 1 and q I 2 results in at least two factors of I 3 , I 4 , I 5 which q divides, as desired.
Suppose q I 1 and q I 2 . We want to show that I 4 cannot be the only factor divisible by q. Suppose then q I 1 and q I 2 , q | I 4 . We show that q divides another factor. For the sake of contradiction, suppose q only divides I 4 .
This completes the proof of the proposition.

Groups of order p 2 q
Next we consider groups of order p 2 q, p, q two distinct primes. We are interested in seeing how far the techniques developed for the case pq can be extended for the case p 2 q. Unfortunately we cannot derive a complete generalization of Proposition 3 and show that DFZ inequalities necessarily hold for these groups. However, we are able to use similar reasoning to derive partial results, which can be used to reduce computation when looking for violators of DFZ inequalities. The case |G 1 | = |G 2 | = p remains open.
The following lemma is an exercise in elementary group theory.
Lemma 8. Let G be a group of order p 2 q. Then G has at least one normal Sylow subgroup.
Proof. Case q < p. We show that the Sylow p-subgroup is normal. Sylow theorems give us the following constraints on the number n p of Sylow p-subgroups: • n p | p 2 q. Together we have n p | q. But as by assumption q < p, the first constraint rules out the case that n p = q. We conclude that n p = 1 so the Sylow p-subgroup is unique. Equivalently, Sylow p-subgroup is normal since Sylow subgroups are conjugates of one another. Case p < q. We have n q = 1 + kq divides p 2 and n p = 1 + lp divides q, k, l ≥ 0, thus n q ∈ {1, p, p 2 }.
• Case n q = 1. Then the Sylow q subgroup is unique, and we are done.
• Case n q = p cannot occur for 1 + kq = p will contradict p < q.
• Case n q = p 2 . We show that this forces n p = 1, i.e. the Sylow p-subgroup to be unique. Note that the number of elements of order q is (q − 1)p 2 , thus the number of elements of order p t , t = 0, 1, 2, is p 2 q − (q − 1)p 2 = p 2 . But that is precisely the order of any Sylow p-subgroup. Therefore all elements of order p are included in a single Sylow p-subgroup. Hence there is a unique Sylow p-subgroup and we are done.
We have concluded that G has at least one unique Sylow subgroup -and it can correspond to either of the prime divisors of G.
As per Remark 1, we will try to take advantage of the uniqueness of this Sylow subgroup and derive a generalization of Proposition 3. We note immediately that if both the Sylow q-subgroup and the Sylow p-subgroup are unique, then G is abelian, and by Corollary 1 we know that the 10 DFZ inequalities (2)-(11) hold. 4.1. Case 1: |G| = p 2 q and the Sylow q-subgroup is normal.
Claim 5. If p > q, then G is in fact abelian. Hence the DFZ inequalities hold for G.
Proof. The Sylow q-subgroup is by assumption normal. It suffices to show that a Sylow p-subgroup is normal, for then G is isomorphic to the product of its (abelian) Sylow subgroups, and hence is itself abelian. But a Sylow p-subgroup G p is of order p 2 , so it has index [G : G p ] = q, which is the smallest prime that divides the order of G. It follows that G p is normal in G.
We could therefore assume that p < q, but we will state our results in full generality.
Lemma 9. Let G be a group of order p 2 q with unique Sylow q-subgroup.
If |G ac | or |G bc | takes values in {1, p 2 q, q, pq}, then G ac G bc is a subgroup. The case when G ac G bc is not a subgroup may then only happen when (|G ac |, |G bc |) takes value in {(p, p), (p, p 2 ), (p 2 , p 2 )}.
Proof. By Claim 5 when p > q the group G is abelian, hence G ac G bc is always a subgroup.
We may therefore assume p < q. In order to show that G ac G bc forms a subgroup, it suffices to show that one of G ac , G bc is normal. We recall that a Sylow subgroup is unique if and only if it is normal.
We will show that any subgroup H of order 1, q, pq, p 2 q is normal. Orders 1, p 2 q follow trivially. Suppose H has order q. By assumption Sylow q-subgroup is unique, so H is the Sylow q-subgroup, and hence H is normal. Hence G ac G bc forms a subgroup.
Suppose H has order pq. Then it is a subgroup of order pq, which has index p, the smallest prime that divides |G|. Thus it is normal. Now recall that for any two subgroups K 1 , K 2 of G, the set K 1 K 2 forms a subgroup of G if either K 1 or K 2 is normal. We have so far shown that if H has order 1, q, pq, p 2 q then it is normal. Hence if either |G ac | or |G bc | takes values in {1, p 2 q, q, pq}, then G ac G bc forms a subgroup. The orders of subgroups of G take value among divisors of p 2 q, these are {1, p, q, p 2 , pq, p 2 q}. By removing {1, p 2 q, q, pq} from the list {1, p, q, p 2 , pq, p 2 q} of possible orders for G ac and G bc , we see that the only orders that might result in G ac G bc not forming a subgroup are {p, p 2 }. The second claim of the lemma now follows.
Proposition 4. Let G be a group of order p 2 q with a normal Sylow q-subgroup G q . Then any two Sylow p-subgroups intersect in a subgroup of order p.
Proof. It is enough to show that G contains a normal subgroup of order p. Indeed, let N be a normal subgroup of order p, it is contained in some Sylow p-group G p . Let G p be another Sylow p-subgroup, which is therefore a conjugate of G p , but since the conjugate of N is N itself, it is contained in G p .
We are left to show that G contains a normal subgroup of order p which is normal.
Since G q is normal in G we have the exact sequence where we use G/G q ∼ = G p . By elementary group theory we know that G p acts on G q by conjugation. In other words, there is a homomorphism from G p to the automorphism group of G q . Case 1: If G p is cyclic of order p 2 then it has a unique subgroup of elements of order p. But that means that this subgroup is normal in G because the conjugate of any element of order p is another element of order p. Case 2: G p = C p × C p . Now Aut(G q ) ∼ = C q−1 is cyclic. Hence α has a kernel K ≤ G p . But then K acts trivially on G q , in other words K is normal in G.
The following proposition is analogous to Proposition 2: it examines the value |G abc ||Gc| |Gac||G bc | , in addition classifying when this value is q p .
Proof. The fact that |G abc ||Gc| |Gac||G bc | ≥ 1 follows from the fact that G ac G bc is a subset of G c . Then Indeed, the restrictions on powers of q follow from Proposition 1 since the Sylow q-subgroup is abelian, while the restrictions on powers of p come from the inclusions G ac ≤ G c , G bc ≤ G c and the fact that v p (G c ) ∈ {0, 1, 2}. The cases |Gc| |GacG bc | = 1 p < 1 and |Gc| |GacG bc | = 1 p 2 < 1 cannot happen. By Claim 5, the case p > q is abelian, thus the result follows. We may therefore assume p < q.
We are left to exclude the case q p 2 , and to characterize when q p occurs. Since p cannot possibly divide q, these ratios appear only when G ac G bc is not a subgroup of G c . By Lemma 9, this may happen when (|G ac |, |G bc |) is (p, p), (p, p 2 ) or (p 2 , p 2 ).
• (p, p) |G abc ||Gc| |Gac||G bc | = 1·|Gc| p·p . Since G ac , G bc are subgroups of G c , we have v p (G c ) ≥ 1, and to obtain |G abc ||Gc| |Gac||G bc | = q/p or q/p 2 , we need v q (G c ) = 1. If |G c | = pq, Since G bc is a subgroup of G c , we have v p (G c ) = 2, and to obtain q/p or q/p 2 , For the same reason as above, G c = G. if |G abc | = 1, |G abc ||Gc| |Gac||G bc | = 1·p 2 q p 2 ·p 2 = q/p 2 , if |G abc | = p, |G abc ||Gc| |Gac||G bc | = p·p 2 q p 2 ·p 2 = q p . The ratio q/p 2 could happen only if the intersection of two Sylow p-subgroups were trivial, a case that never occurs, by Proposition 4.
The next proposition is a partial generalization to groups of order p 2 q of Proposition 3 on groups of order pq. Note that reducing to the case |G 1 | = |G 2 | = p can be used to reduce computation when looking for violators of DFZ inequalities.
Evaluating |G12||G| |G1||G2| in the LHS of the 10 DFZ inequalities (2)- (11) we show that we only need to consider the case when the LHS is q p : • (p, p) note that this case is excluded by assumption if we are done since G 1 has order p, and it intersects G 2 in a subgroup of order p, implying that G 1 is a subset of G 2 , and inequalities hold by Corollary 1.
-|G 12 | = 1 cannot happen since no two Sylow p-subgroups intersect trivially by Proposition 4. - We have so far shown that the cases when G 1 G 2 does not form a subgroup will have the LHS of each inequality corresponding to q p , unless |G 1 | = |G 2 | = p. Claim 6. Suppose that the LHS is q p . If v q (RHS) ≥ 1, then the inequalities hold.
Terms on the right hand side can only take values among {1, p, q, pq, p 2 , p 2 q, q p } by Proposition 5. The assumption v q (RHS) ≥ 1 means that at least one term on the RHS is taking values among {q, pq, pq 2 , q p }. Since each of these is ≥ q p , while all other terms on the RHS are ≥ 1, we conclude LHS ≤ RHS and we have proved the claim.
To complete the proof we must show that indeed v q (RHS) ≥ 1. This has been shown for cases of groups of order pq and the proofs are identical.
Let us comment on the case (|G 1 |, |G 2 |) = (p, p). Then |G12||G| |G1||G2| = 1·(p 2 q) p·p = q and the proof technique used above relying on v q (RHS) ≥ 1 is not sufficient anymore. Say G 1 , G 2 have orders (p, p) so the LHS of an inequality is q. But on the RHS we may have q/p. By Proposition 5, this will exactly happen when An inequality of the form q ≤ q/p · I 1 · I 2 · I 3 where q/p corresponds e.g. to |G15||G|

|G1||G5|
does not have to be true. It depends on the terms I, I 2 , I 3 , therefore the techniques developed in this paper do not apply immediately to this case.
Example 1. As a concrete example, consider the dihedral group D 20 = r, s, r 10 = s 2 = 1 . There is one Sylow 5-subgroup, namely r 2 . There are five Sylow 2subgroups, given by s, r 5 = {1, s, r 5 , sr 5 } and its four conjugates, for example r s, r 5 r −1 = {1, rsr −1 , r 5 , rsr 4 }. We notice the subgroup {1, r 5 } which is in the intersection of both these Sylow 2-subgroups. Take Lemma 10. Let G be a group of order p 2 q whose Sylow p-subgroup is normal. Then the quantity i a;b|c ≥ 1 and i a;b|c ∈ {1, q, p, pq, p 2 , p 2 q, p q , p 2 q }.
Proof. The fact that i a;b|c ≥ 1 follows from the fact that i a;b|c = |G abc ||Gc| |Gac||G bc | = |Gc| |GacG bc | and G ac G bc is a subset of G c , as before. Next The arguments is as before, the restrictions on powers of p follow from Proposition 1 since the Sylow p-subgroup is abelian, while the restrictions on powers of q come from the inclusions G ac ≤ G c , G bc ≤ G c and the fact that v q (G c ) = 0, 1. The case 1/q < 1 cannot happen.
Next we look at the case when |G| = p 2 q, G p is normal, while G q is not. If p < q, then, as shown below, the group is forced to be A 4 .
Claim 7. Suppose |G| = p 2 q, G p is normal, while G q is not, and p < q. Then G ∼ = A 4 . Hence 10 DFZ inequalities hold for G Proof. Since G p is normal and G q ∼ = C q , we have an exact sequence 1 → G p → G → C q → 1 and C q acts on G p , i.e. there is a homomorphism Note that the assumption that G q is not normal implies that G cannot be abelian. This means that the action induced by α is nontrivial.
There are two cases to consider: • If G p = C p 2 , then G would in fact be forced to be abelian: We recall that the automorphism group Aut(C p 2 ) ∼ = Z × p 2 ∼ = Z p−1 × Z p has order (p − 1)p. Hence the image of any homomorphism α : C q → Aut(G p ) divides p − 1. But it must also divide |C q | = q. Since by assumption p < q, we conclude that the image of α is {1}, i.e., α is trivial. In other words α induces trivial action of G q on G p . This makes G an abelian group, contradicting the assumption that G q is not normal.
• If G p = C p ×C p , we know that Aut(G p ) is the general linear group GL 2 (F p ) of order (p 2 − 1)(p 2 − p). We consider the case when the image of α is nontrivial, so is of order q. For q to divide (p 2 − 1)(p 2 − p), together with the assumption p < q, we would need that q | p + 1 and so p = 2 and q = 3. Therefore G is a nonabelian semidirect product of C 2 × C 2 and C 3 . Up to isomorphism the only such group is A 4 , as claimed.
We have completed showing the claim. It can be checked numerically that A 4 does not violate any of the 10 DFZ inequalities.
Therefore we may assume that q < p. Here is the analogue of Lemma 9.
Lemma 11. Let G be a group of order p 2 q with normal Sylow p-subgroup. Assume that q < p. If |G ac | or |G bc | takes values in {1, p 2 q, p, p 2 }, then G ac G bc is a subgroup. The case when the set G ac G bc is not a subgroup happens only when the orders (|G ac |, |G bc |) take values in {(q, q), (q, pq), (pq, pq)}.
Proof. It suffices to show that any subgroup H of order 1, p 2 q, p, p 2 is normal.
Orders 1, p 2 q follow trivially. For order p, it follows from the uniqueness of the Sylow p-subgroup. For p 2 , this is because a subgroup of order p 2 has index q, the smallest prime that divides |G|, thus it is normal. The second claim follows from removing {1, p 2 q, p, p 2 } from the list {1, q, p, p 2 , pq, p 2 q} of possible orders.
Therefore, we may conclude as in Proposition 6.
Proposition 7. Let G be a group of order p 2 q with normal Sylow p-subgroup. Let G 1 , G 2 , G 3 , G 4 , G 5 be arbitrary subgroups of G. The 10 DFZ inequalities hold for Proof. The case p < q is taken care of by Claim 7. The proof for the case q < p, where we now apply Lemma 11 instead of Lemma 9 is similar as in Proposition 6. As a result of Lemma 11, the only non-normal candidates for G 1 , G 2 will give as possible values for (|G 1 |, |G 2 |) the pairs (q, q), (q, pq), (pq, pq).
We observe that the cases for which it remains to establish the inequality have v p (LHS) ≥ 1. We proceed with the proof as in Proposition 5, only now we use valuation v p at p in an analogue of Claim 6.

Group violations
5.1. Violations by S 4 . We prove that the symmetric group S 4 violates two of the ten DFZ inequalities, and so does the projective group P GL 2 (F q ), for q a prime power.
Let GL 2 (F p ) be the general linear group of order 2 over F p . It has order (p 2 − 1)(p 2 − p), and naturally acts on the projective line P(F p ), whose p + 1 elements can be written (in homogeneous coordinates) as {(0, 1), (1, 1), . . . , (p − 1, 1), (1, 0)}. This action descends to a faithful action of P GL 2 (F p ), the projective general linear group over F p of degree 2, on P(F p ), which permutes the elements of P(F p ). This thus yields an injective homomorphism from P GL 2 (F p ) to the symmetric group S p+1 . For p = 3, |P GL 2 (F 3 )| = 24 = |S 4 |, showing that P GL 2 (F p ) S 4 . This suggests to look at the group P GL 2 (F q ), q a prime power, to find examples of violators. Indeed, it is exactly known when S 4 is contained in P GL 2 (K), for K an arbitrary field.
Proposition 9 ([13, 2.5]). Suppose that the characteristic of K is prime to the order of S 4 . Then P GL 2 (K) contains S 4 if and only if −1 is a sum of two squares in K.
It follows that whenever p ≥ 5, then P GL 2 (F q ) contains S 4 , for q = p r , for r ≥ 1. The proof of the following result is attributed to Henry Mann [8] (the result holds for characteristic 2, with a different proof).
Proposition 10. Suppose that the characteristic of F q is not 2. Every element of F q is a sum of two squares.
Proof. Consider the map φ, which maps x to x 2 . Suppose φ(a) = φ(b), then a 2 = b 2 ⇐⇒ (a − b)(a + b) = 0 ⇐⇒ a = b or a = −b. Therefore φ restricted to F * q sends two distinct elements of F * q to a single element, from which we deduce that exactly half of the elements of F * q are squares. Since 0 is also a square, the set S of squares in F q contains (q + 1)/2 elements.
Next, we look at the additive structure of F q , and pick a ∈ F q . Notice that |S| = |a − S| = |{a − s, s ∈ S}|. Since |S| + |a − S| > |F q |, |S| ∩ |a − S| is not empty. Take s = a − s in the intersection, then a = s + s, and is therefore a sum of two squares.
Therefore another example of group violating the inequalities (2) and (4) after S 4 is P GL 2 (F 5 ) of order 120, which turns out to be isomorphic to S 5 (the same proof as above holds).

Smallest violators.
From Corollary 1, abelian groups never violate the 10 DFZ inequalities. From Proposition 3, neither do groups of order pq for p, q two distinct primes. Thus until order 23 (included), only 8, 12, 16, 18, and 20 are orders where potential violators could exist. Suppose there exists an abelian group A such that for every choice of subgroups G 1 , . . . , G n of G, there exist subgroups A 1 , . . . , A n of A such that [G : G A ] = [A : A A ] for every non-empty subset A of {1, . . . , n}, and G A = ∩ i∈A G i , A A = ∩ i∈A A i . Then we say that G is abelian group representable. Groups of order 8 are known to be abelian group representable [14] and can be ruled out as well. Groups of orders 12,16,18,20 are left, apart 16, all of them fit in the category of groups of order p 2 q. However, since the case of |G 1 | = |G 2 | = p is still incomplete, we checked numerically that no violation of the 10 DFZ inequalities is to be found. Therefore the smallest violator is of order 24. 5.3. Simultaneous violators. One may wonder whether it is possible to find a simultaneous group violator for the ten DFZ inequalities. We provide a negative answer. Definition 1. If a finite group G and subgroups G 1 , . . . , G n violate two or more inequalities, then (G, G 1 , . . . , G n ) is called a simultaneous group violator for those inequalities.
Proposition 11. There do not exist any simultaneous group violators for (2) and(4).
Proof. Suppose (G, G 1 , G 2 , G 3 , G 4 , G 5 ) are a simultaneous group violator of the group inequalities of (2) and (4). That is,  Proof. If there exists a simultaneous group violator for all DFZ inequalities, it violates (2) and (4) simultaneously, a contradiction to Proposition 11.