$p$-groups with exactly four codegrees

TL;DR

This paper studies $p$-groups with exactly four codegrees, establishing bounds on their nilpotence class and order under various conditions, extending previous results known for groups with fewer codegrees.

Contribution

It extends the classification of $p$-groups with three codegrees to those with four, providing bounds on class and order under specific structural hypotheses.

Findings

If $G$ has two irreducible character degrees, its nilpotence class is at most 4.

For groups with coclass at most 3, the order is bounded by $p^7$.

With additional hypotheses, the order is bounded by $p^{11}$ for coclass at most 7.

Abstract

Let $G$ be a $p$-group and let $\chi$ be an irreducible character of $G$. The codegree of $\chi$ is given by $|G:\text{ker}(\chi)|/\chi(1)$. Du and Lewis have shown that a $p$-group with exactly three codegrees has nilpotence class at most 2. Here we investigate $p$-groups with exactly four codegrees. If, in addition to having exactly four codegrees, $G$ has two irreducible character degrees, $G$ has largest irreducible character degree $p^2$, $|G:G'|=p^2$, or $G$ has coclass at most 3, then $G$ has nilpotence class at most 4. In the case of coclass at most 3, the order of $G$ is bounded by $p^7$. With an additional hypothesis we can extend this result to $p$-groups with four codegrees and coclass at most 7. In this case the order of $G$ is bounded by $p^{11}$.

Full text

1. Introduction

In this paper all groups are finite $p$-groups. For an irreducible character $\chi$ of a group $G$, the codegree of $\chi$ is defined as $\text{cod}(\chi)=|G:\text{ker}(\chi)|/\chi(1)$. The set of irreducible characters of $G$ is denoted $\text{Irr}(G)$, the set of codegrees of the irreducible characters of $G$ is denoted $\text{cod}(G)$, and the nilpotence class of $G$ is $c(G)$. In [4], Du and Lewis showed that a $p$-group with exactly two codegrees is elementary abelian, and a $p$-group with exactly three codegrees has nilpotence class at most 2. Our original purpose was to find a sharp bound for the nilpotence class of groups with exactly four codegrees. Ideally, this bound would be a simple constant. At this time, our best result for arbitrary $p$-groups depends on the largest codegree, but in several more specific cases we have attained the predicted bound of nilpotence class at most 4.

Theorem 1.1**.**

Let $G$ be a finite $p$-group such that $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If any of the following hold, then $G$ has nilpotence class at most 4:

1. (i)

$|\text{cd}(G)|=2$, 2. (ii)

$\text{cd}(G)=\{1,p,p^{2}\}$, 3. (iii)

$|G:G^{\prime}|=p^{2}$.

The coclass of a $p$-group $G$ with nilpotence class $c$ is given by $\log_{p}{|G|}-c$. If $G$ has four codegrees and coclass at most $3$, then $G$ has nilpotence class at most 4, bounding the order of $G$.

Theorem 1.2**.**

Let $G$ be a $p$-group such that $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$, where $2\leq b<a$. If $G$ has coclass at most 3, then $G$ has nilpotence class at most $4$, and $|G|\leq p^{7}$.

With the following additional hypothesis, we can extend the result of Theorem 1.2 to $p$-groups with coclass at most $7$.

Hypothesis $(\ast)$.

If $G$ is a $p$-group with nilpotence class $n$ such that $|G|\geq p^{2n}$, then $|Z_{2}(G)|\neq p^{2}$.

Theorem 1.3**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has coclass at most 7, and $G$ and all of its quotients satisfy Hypothesis $(\ast)$, then the nilpotence class of $G$ is at most 4 and $|G|\leq p^{11}$.

In [4], Du and Lewis were able to bound $c(G)$ in terms of the largest member of $\text{cod}(G)$. They showed that if $p^{a}$ (where $a>1$) is the largest codegree of $G$, then $c(G)\leq 2a-2$, and in some specific cases, $c(G)\leq 2a-3$. When $|\text{cod}(G)|=4$, we can improve this bound.

Theorem 1.4**.**

If $G$ is a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$, then $c(G)\leq a+1$.

This bound can be improved slightly when the two largest codegrees are consecutive powers of $p$ and the group does not have $p^{2}$ as a codegree.

Theorem 1.5**.**

If $G$ is a finite $p$-group such that $\text{cod}(G)=\{1,p,p^{a-1},p^{a}\}$ for $a\geq 4$, then ${c}(G)\leq a$.

We expect this work to appear as part of the first author’s Ph.D. dissertation at Kent State University.

2. Bounding nilpotence class

Our first lemma relates the codegree of a faithful irreducible character to the order of the group.

Lemma 2.1**.**

[4, Lemma 3.1]** Let $G$ be a group and suppose that $\chi\in\text{Irr}(G)$ is faithful. Then $|G|=\chi(1)\text{cod}(\chi)<\text{cod}(\chi)^{2}$.

Lemma 2.2 can be inferred from [4].

Lemma 2.2**.**

Let $G$ be a $p$-group with $p^{2}\notin\text{cod}(G)$. If $\chi$ is an irreducible character of $G$ such that $\text{cod}(\chi)>p$, then $\chi$ is non-linear.

Proof.

Let $\chi$ be an irreducible character of $G$ such that $\text{cod}(\chi)=p^{a}$ for some $a>2$, and suppose that $\chi$ is linear. Then $\text{ker}(\chi)\geq G^{\prime}$, so $G/\text{ker}(\chi)$ is abelian. Since $\chi$ is a faithful irreducible character of $G/\text{ker}(\chi)$, this quotient must be cyclic, and $|G/\text{ker}(\chi)|=\chi(1)\text{cod}(\chi)=p^{a}>p^{2}$. By Corollary 2.5 of [4], $G/G^{\prime}$ is elementary abelian, which is impossible since $\text{ker}(\chi)\geq G^{\prime}$ and $G/\text{ker}(\chi)$ is cyclic with order greater than $p^{2}$. ∎

The following useful lemma is a consequence of Proposition 2.5 of [1] and Itô’s Theorem [6, Theorem 6.15].

Lemma 2.3**.**

If a finite $p$-group $G$ has a faithful irreducible character of degree $p$, then $G$ has a normal abelian subgroup of index $p$ and $\text{cd}(G)=\{1,p\}$.

A group with $\text{cd}(G)=\{1,p\}$ must satisfy one of two conditions first given by Isaacs and Passman in [7] as Theorem C 4.8. We state this theorem below as it appears in [1].

Lemma 2.4**.**

[1*, Theorem 22.5]**

If $G$ is a nonabelian $p$-group with $\text{cd}(G)=\{1,p\}$, then one and only one of the following holds:*

1. (i)

$G$* has an abelian subgroup of index $p$,* 2. (ii)

$G/Z(G)$* is of order $p^{3}$ and exponent $p$.*

Frequently, we will show through Lemma 2.3 or Lemma 2.4 that a group has an abelian subgroup of index $p$, and then use the following lemma to obtain a contradiction about the possible orders of the group, the derived subgroup, and the center.

Lemma 2.5**.**

[1, Lemma 1.1]** Let $A$ be an abelian subgroup of index $p$ of a nonabelian $p$-group $G$. Then $|G|=p|G^{\prime}||Z(G)|$.

The proofs of the main theorems will make use of minimal counter examples to reach a contradiction. The next lemma will allow us to shorten these proofs.

Lemma 2.6**.**

Let $G$ be a finite $p$-group such that $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$, where $2\leq b<a$. If $|G|$ is minimal such that $G$ has nilpotence class $n\geq 3$, then $G$ has a faithful irreducible character.

Proof.

Let $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ and let $|G|$ be minimal such that $G$ has class $n$. Let $K$ be the kernel of an irreducible character of $G$. Either $\text{cod}(G/K)\subsetneq\text{cod}(G)$, in which case $c(G/K)\leq 2$, or $\text{cod}(G/K)=\text{cod}(G)$, and since $|G|$ is minimal, $c(G/K)\leq n-1$. Hence, there is a central series of $G$ with the $n-1$ term contained in $K$ for all kernels of irreducible characters of $G$, and since the intersection of these kernels is trivial, $G$ has class at most $n-1$, a contradiction. Thus $G$ must have a faithful irreducible character.∎

The upper bound on the nilpotence class of $G$ when $p^{a}=\text{max}(\text{cod}(G))$ is given as $2a-2$ or $2a-3$ in [4]. Restricting $|\text{cod}(G)|$ to 4 yields the stronger bound of Theorem 1.4.

Proof of Theorem 1.4.

Let $G$ be a minimal counterexample and notice that $c(G)=a+2$, for otherwise $c(G)\geq a+3$ implies $c(G/Z)\geq a+2\geq 5$, and by Theorem 1.2 of [4], $\text{cod}(G/Z)=\{1,p,p^{b},p^{a}\}$, contradicting that $|G|$ is minimal. By Lemma 2.6, $G$ has a faithful character of codegree at most $p^{a}$ and by Lemma 2.1, $|G|\leq p^{2a-1}$. Using Theorem 1.2 of [4] and that $c(G/Z_{a-1})=3$, we have $\text{cod}(G/Z_{a-1})=\text{cod}(G)$. Let $\chi$ be an irreducible character of $G/Z_{a-1}$ such that $\text{cod}(\chi)=p^{a}$. Then $p^{a}\leq\chi(1)p^{a}=|G:\text{ker}(\chi)|\leq|G:Z_{a-1}|\leq p^{a}$ so $\chi$ is a faithful linear character of $G/Z_{a-1}$, which is impossible since $G/Z_{a-1}$ has class 3. Hence $c(G)\leq a+1$. ∎

Theorem 1.5 further improves the bound in Theorem 1.4 when the two largest codegrees are consecutive powers of $p$ and $p^{2}\notin\text{cod}(G)$.

Proof of Theorem 1.5.

Let $G$ be a minimal counterexample. Then ${c}(G)=a+1$ and by Lemma 2.6, $G$ has a faithful character with codegree at most $p^{a}$, giving $|G|\leq p^{2a-1}$. The quotient $G/Z_{a-2}$ has class 3 and therefore has the same set of codegrees as $G$. Let $\chi$ be an irreducible character of $G/Z_{a-2}$ such that $\text{cod}(\chi)=p^{a}$. By Lemma 2.2, $\chi$ is nonlinear so we have $p^{a+1}\leq\chi(1)p^{a}=|G:\text{ker}(\chi)|\leq|G:Z_{a-2}|\leq p^{a+1}$, which shows that $\chi$ is faithful, $\chi(1)=p$ and by Lemma 2.3, $\text{cd}(G/Z_{a-2})=\{1,p\}$. Let $\mu$ be an irreducible character of $G/Z_{a-2}$ with codegree $p^{a-1}$. Then $p^{a}=\mu(1)\text{cod}(\mu)=|G:\text{ker}(\mu)|\leq|G:Z_{a-2}|=p^{a+1}$, which shows that $|\text{ker}(\mu):Z_{a-2}|=p$ and hence $\text{ker}(\mu)\leq Z_{a-1}$.

Suppose $|Z_{a-1}:Z_{a-2}|=p$. Then $\mu$ is a faithful character of $G/Z_{a-1}$, and $Z_{a}/Z_{a-1}$ is cyclic, so $Z_{a-1}\leq G^{\prime}\leq Z_{a}$. Since $p^{2}\notin\text{cod}(G)$, we have that $G/G^{\prime}$ is elementary abelian and hence $|Z_{a}:G^{\prime}|\leq p$. Notice that $Z(\mu)=Z_{a}$, and by Lemma 2.31 of [6], $|G:Z(\mu)|=p^{2}$. We also have $|G^{\prime}:Z_{a-1}|=p$, as $G^{\prime}\geq Z_{a-1}\geq G_{3}$ and $G^{\prime}/G_{3}$ is elementary abelian while $G^{\prime}/Z_{a-1}$ is cyclic. If $G^{\prime}=Z_{a}$, then $|G/Z_{a-2}|=p^{4}$, which contradicts that $\chi$ is a faithful nonlinear character with codegree $p^{a}\geq p^{4}$. Thus $|Z_{a}:G^{\prime}|=p$ and we have $|G/Z_{a-2}|=p^{5}$. By Lemma 2.6 of [3], this contradicts $p^{2}\notin\text{cod}(G)$.

We may now assume that $|Z_{a-1}:Z_{a-2}|>p$. Let $G/Z_{a-2}=H$, so $H$ has class 3, $|Z(H)|>p$, and $\mu$ is an irreducible character of $H$ with $|\text{ker}(\mu)|=p$. Since $\chi$ is a faithful irreducible character of $H$, $Z(H)$ is cyclic and hence $\text{ker}(\mu)$ is the unique subgroup of $H$ of order $p$. Notice that $H_{3}$ is elementary abelian, and since it is contained in the cyclic subgroup $Z(H)$, we have $H_{3}=\text{ker}(\mu)$. As $H/H_{3}$ has class 2, $H^{\prime}/H_{3}\leq Z(H/H_{3})=Z(\mu)/H_{3}$, which shows that $H/Z(\mu)$ is abelian and hence $|H:Z(\mu)|=p^{2}$. We then have that $Z(\mu)=Z_{2}(H)$, as $Z(\mu)\leq Z_{2}(H)$. By Lemma 2.27 of [6], $Z(\mu)/\text{ker}(\mu)=Z_{2}(H)/H_{3}$ is cyclic. We also have $H_{3}\leq H^{\prime}\leq Z_{2}(H)$, so $H^{\prime}/H_{3}$ is cyclic as well as elementary abelian, and hence $|H^{\prime}:H_{3}|=p$. This is impossible since $|Z(H):H_{3}|=p$ and $|H^{\prime}:Z(H)|\geq p$. Thus $c(G)\leq a$. ∎

The next lemma can be used to further shorten our proofs bounding the nilpotence class when a group has exactly four codegrees. With this lemma, we can assume that such a group with nilpotence class greater than 4 has at least one faithful irreducible character, and the codegree of any such character must be as large as possible.

Lemma 2.7**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has nilpotence class $5$, then faithful irreducible characters of $G$ have codegree $p^{a}$.

Proof.

Let $\chi\in\text{Irr}(G)$ be faithful and suppose $\text{cod}(\chi)=p^{b}$. If $b=2$, then $|G|\leq p^{5}$ by Lemma 2.1, which is impossible since $G$ has nilpotence class 5. Hence $b\geq 3$ and $p^{2}\notin\text{cod}(G)$. Let $\varphi\in\text{Irr}(G/Z_{3})$ be nonlinear, and let $\text{cod}(\varphi)=p^{r}$. Note that since $\varphi$ is nonlinear, $r$ is equal to either $b$ or $a$. Put $\text{ker}(\varphi)=K$ and $Z(\varphi)=Y$. By definition, $\varphi(1)p^{r}=|G:K|$, and since $G/Z_{3}$ has nilpotence class 2, $|G:Y|=\varphi(1)^{2}$ by Theorem 2.31 of [6]. Combining these equations yields $p^{r}\varphi(1)=\varphi(1)^{2}|Y:K|$, and hence $p^{r}=\varphi(1)|Y:K|$. Now $|G|=\chi(1)p^{b}\leq\chi(1)p^{r}=\chi(1)\varphi(1)|Y:K|$, so $\chi(1)\varphi(1)\geq|G:Y||K|=\varphi(1)^{2}|K|$. This shows that $p^{r}>\chi(1)\geq\varphi(1)|K|$, so $\varphi(1)|K|<p^{r}=\varphi(1)|Y:K|$. Finally, this implies $|K|<|Y:K|$.

Since $G/K$ has class 2, $G/Y$ is abelian, and hence $G^{\prime}\leq Y$. Thus $Y\geq G^{\prime}K\geq K$, and since $Y/K$ is cyclic, the quotient $Y/G^{\prime}K$ is also cyclic. Recalling that $G/G^{\prime}$ is elementary abelian since $p^{2}\notin\text{cod}(G)$, this also shows that $Y/G^{\prime}K$ is elementary abelian. Being both cyclic and elementary abelian, the order of $Y/G^{\prime}K$ is at most $p$. Using the fact that $Y$ contains $G^{\prime}$, we have $G_{3}=[G^{\prime},G]\leq[Y,G]\leq K$. This shows that $G^{\prime}\cap K$ contains $G_{3}$. Since $G^{\prime}/G_{3}$ is elementary abelian, $G^{\prime}/G^{\prime}\cap K$ is also elementary abelian. By the Diamond Isomorphism Theorem, $G^{\prime}/G^{\prime}\cap K$ is isomorphic to $G^{\prime}K/K$, which is then elementary abelian, but also cyclic, since $Y/K$ is cyclic. This shows that the order of $G^{\prime}K/K$ is at most $p$. We now have $|Y:K|=|Y:G^{\prime}K||G^{\prime}K:K|\leq p^{2}$. Since $\varphi$ is an irreducible character of $G/Z_{3}$, we know that $p^{3}\leq|Z_{3}|\leq|K|$, but the previous paragraph showed that $|K|<|Y:K|$, which is at most $p^{2}$, a contradiction. Hence, any faithful characters of $G$ must have codegree $p^{a}$. ∎

Recall for a $p$-group with exactly two character degrees that there is no bound on the nilpotence class of $G$ when $\text{cd}(G)=\{1,p\}$, and otherwise $G$ has nilpotence class at most $p$. If the group also has exactly four codegrees, then it has nilpotence class at most 4.

Proof of Theorem 1.1 (i).

Let $G$ be a minimal counterexample. By Lemmas 2.6 and 2.7, $G$ has at least one faithful irreducible character $\chi$, and such a character must have codegree $p^{a}$. Let $|G|=p^{n}$, and notice that $\chi(1)=p^{n-a}$. Since $c(G/Z)=4$, Theorem 1.2 of [4] implies that $\text{cod}(G/Z)=\text{cod}(G)$. Let $\varphi\in\text{Irr}(G/Z)$ have codegree $p^{a}$. If $\varphi$ is nonlinear, then $\varphi(1)=p^{n-a}$, and hence $|G:\text{ker}(\varphi)|=\text{cod}(\varphi)\varphi(1)=p^{a}p^{n-a}=p^{n}=|G|$. This shows that $\varphi$ is a faithful irreducible character of $G$, contradicting that the kernel of $\varphi$ contains $Z$. Hence $\varphi$ must be linear. By Lemma 2.2, $p^{2}\in\text{cod}(G)$, and by a similar argument, $p^{a}=p^{3}$. By Lemma 2.1, we have $|G|<\text{cod}(\chi)^{2}=p^{6}$, implying that $|G|\leq p^{5}$, and hence $G$ has nilpotence class at most 4. ∎

Proof of Theorem 1.1 (ii).

Let $G$ be a minimal counterexample. Notice that for the kernel $K$ of an irreducible character of $G$, if $\text{cod}(G/K)=\text{cod}(G)$ then $\text{cd}(G/K)=\text{cd}(G)$, for otherwise we have a contradiction with Theorem 1.1 (i). Since $|G|$ is minimal, we must have $c(G/K)<c(G)$, and we can apply Lemmas 2.6 and 2.7. Now $c(G)=5$ and $G$ has a faithful irreducible character $\chi$ with codegree $p^{a}$. By Lemma 2.3, $\chi$ must have degree $p^{2}$, for otherwise $|\text{cd}(G)|=\{1,p\}$, a contradiction. This implies that $p^{a+2}=\chi(1)\text{cod}(\chi)=|G|$, and since $G$ has class 5, we see that $a$ is at least 4.

The quotient $G/Z_{2}$ has nilpotence class 3, hence $\text{cod}(G/Z_{2})=\text{cod}(G)$ and we can find $\gamma\in\text{Irr}(G/Z_{2})$ with codegree $p^{a}$. Since $a\geq 4$ and $|\text{cod}(G)|=4$, either $p^{2}$ or $p^{3}$ is not in $\text{cod}(G)$. Hence $\gamma$ is nonlinear, and we have $p^{a+1}\leq\gamma(1)\text{cod}(\gamma)\leq|G:Z_{2}|\leq p^{a}$, which is impossible. Thus $G$ can have nilpotence class at most 4. ∎

To complete the proof of Theorem 1.1, we will use the following lemma for maximal class $p$-groups.

Lemma 2.8**.**

[2*, Theorem 2.4]**

If $G$ is a maximal class $p$-group such that $|G|\geq p^{4}$, then $p^{3}\in\text{cod}(G)$.*

When the derived subgroup of a $p$-group $G$ has index $p^{2}$ in $G$, some restrictions are placed on the structure of $G$, including those in the following lemma. In the proof of Theorem 1.1 (iii), we use these facts to show that if such a group has exactly four codegrees, then it must have nilpotence class at most 4.

Lemma 2.9**.**

Let $G$ be a nonabelian $p$-group such that $|G:G^{\prime}|=p^{2}$. Then $|G:G_{3}|=p^{3}$, and $|G:G_{4}|\leq p^{5}$.

Proof.

Since $|G:G^{\prime}|=p^{2}$, $G$ is two-generated, and we can write $G=\langle x,y\rangle$. Then $G^{\prime}/G_{3}=\langle[x,y]\rangle/G_{3}$, and since $G/G^{\prime}$ is elementary abelian, $|G^{\prime}:G_{3}|=p$. For $G_{3}/G_{4}$, we have $G_{3}/G_{4}=\langle[x,y,x],[x,y,y]\rangle/G_{4}$, and hence $|G_{3}:G_{4}|=p$ or $p^{2}$. ∎

Proof of Theorem 1.1 (iii).

Let $G$ be a minimal counterexample. Then $G$ has nilpotence class 5, and by Lemmas 2.6 and 2.7, $G$ has a faithful character $\chi$ with codegree $p^{a}$. Let $\varphi\in\text{Irr}(G/G_{3})$ be nonlinear. As $G/G_{3}$ has class 2, $G/Z(\varphi)$ is abelian and $|G:Z(\varphi)|=\varphi(1)^{2}$. Since $Z(\varphi)$ contains $G^{\prime}$ which has index $p^{2}$ in $G$, we see that $\varphi(1)=p$. Put $\text{cod}(\varphi)=p^{r}$. Then $p^{r+1}=\text{cod}(\varphi)\varphi(1)=|G:\text{ker}(\varphi)|\leq|G:G_{3}|$. By Lemma 2.9, $|G:G_{3}|=p^{3}$, which implies $r=2$ and $\text{cod}(G/G_{3})=\{1,p,p^{2}\}$.

The nilpotence class of $G/G_{4}$ is 3, which implies that $\text{cod}(G/G_{4})=\{1,p,p^{2},p^{a}\}$. Let $\theta\in\text{Irr}(G/G_{4})$ be nonlinear with codegree $p^{a}$. Then $p^{a+1}\leq\text{cod}(\varphi)\varphi(1)=|G:\text{ker}(\varphi)|\leq|G:G_{4}|$. By Lemma 2.9, $|G:G_{4}|\leq p^{5}$, which implies $a\leq 4$. Recalling that $G$ has a faithful character with codegree $p^{a}$, we have that $|G|\leq p^{7}$. If $|G|=p^{6}$ , then by Lemma 2.8, $a=3$, implying that $|G|$ is actually at most $p^{5}$, which contradicts that $G$ has nilpotence class 5. Hence $|G|=p^{7}$ and $a=4$.

We now have $|G:G^{\prime}|=p^{2}$, $|G:G_{3}|=p^{3}$, and $|G:G_{4}|=p^{4}$ or $p^{5}$. If $|G_{5}|=p^{2}$, then $G/G_{5}$ has maximal class, and by Lemma 2.8, $p^{3}\in\text{cod}(G/G_{5})$, a contradiction. Hence $|G_{5}|=p$. Suppose $|G:G_{4}|=p^{4}$ and let $N$ be a normal subgroup of $G$ such that $G_{5}<N<G_{4}$. Then $G/N$ has order $p^{5}$ and class 4, and we can again obtain a contradiction from Lemma 2.8. Now let $|G:G_{4}|=p^{5}$ and let $N$ be a normal subgroup of $G$ such that $G_{4}<N<G_{3}$. Then $G/N$ has order $p^{4}$ and class 3, and Lemma 2.8 yields the desired contradiction. Thus, no minimal counterexample exists. ∎

This proves Theorem 1.2 in the case when $G$ has coclass 1. Notice that coclass 1 is equivalent to having maximal class, and a $p$-group with nilpotence 4 that is also maximal class has order $p^{5}$.

Lemma 2.10**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has coclass 2, then the nilpotence class of $G$ is at most 4, and $|G|\leq p^{6}$.

Proof.

Let $G$ be a minimal counterexample and suppose $c(G)=c>5$. The nilpotence class of $G/Z$ is $c-1>4$, so $G/Z$ must be maximal class since $|G|$ was minimal with coclass 2, but this contradicts Theorem 1.1 (iii). Hence we may assume $G$ has nilpotence class 5.

Suppose $G$ has no faithful irreducible character. If $K$ is the kernel of an irreducible character of $G$, then either $c(G/K)\leq 4$ or $c(G/K)=5$ and $G/K$ has maximal class. Since the latter contradicts Theorem 1.1 (iii), we must have $c(G/K)\leq 4$ for all irreducible characters of $G$ which contradicts that $G$ has class 5. Hence $G$ has a faithful irreducible character $\chi$, and by Lemma 2.7, $\chi$ has codegree $p^{a}$.

As $G$ has class 5 and coclass 2, we see that $|G|=p^{7}$. If $|Z_{2}|=p^{3}$, then $G/Z_{2}$ has class 3 and order $p^{4}$, so it has maximal class, and by Lemma 2.3 of [2] and Lemma 2.8, $\text{cod}(G)=\{1,p,p^{2},p^{3}\}$. Since $G$ has a faithful character $\chi$ with codegree $p^{a}=p^{3}$, by Lemma 2.1, $|G|\leq p^{5}$, a contradiction. Hence $|Z_{2}|=p^{2}$ and $|Z|=p$. By Lemma 2.2 of [3], we have $|Z_{4}|=p^{5}$.

Let $\theta\in\text{Irr}(G/Z_{2})$ have codegree $p^{a}$ and notice that $\theta$ is nonlinear. Thus $p^{a+1}\leq\theta(1)\text{cod}(\theta)\leq|G:Z_{2}|=p^{5}$. Since $a\geq 4$, this shows that $\theta$ is a faithful character of $G/Z_{2}$ and has degree $p$. By Lemma 2.3, $\text{cd}(G/Z_{2})=\{1,p\}$. Now Lemmas 2.4 and 2.5 imply that either $|G:Z_{3}|=p^{3}$ or $p^{5}=|G:Z_{2}|=p|Z_{3}:Z_{2}||G^{\prime}:Z_{2}|$. Suppose $|G:Z_{3}|\neq p^{3}$. Then $|Z_{3}|=p^{3}$ and $|G^{\prime}:Z_{2}|=p^{3}$, contradicting Theorem 1.1 (iii). Hence $|G:Z_{3}|=p^{3}$. By [5], Theorem 2.47, $G^{\prime}>Z_{3}$, which forces $G^{\prime}$ to have order $p^{5}$. This contradicts Theorem 1.1 (iii), and hence $G$ must have nilpotence class at most 4. Let $G$ have order $p^{n}$ and nilpotence class $c\leq 4$. Since $G$ has coclass $2$, we have $n-2=c\leq 4$, and hence $|G|\leq p^{6}$. ∎

This proves Theorem 1.2 in the case when $G$ has coclass 2. The next lemma completes the proof of Theorem 1.2.

Lemma 2.11**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has coclass 3, then the nilpotence class of $G$ is at most 4, and $|G|\leq p^{7}$.

Proof.

Let $G$ be a minimal counterexample and suppose $c(G)=c>5$. The nilpotence class of $G/Z$ is $c-1>4$, so $G/Z$ must be maximal class or have coclass 2 since $|G|$ was minimal with coclass 3, but this contradicts Theorem 1.1 (iii) and Lemma 2.10. Hence we may assume $G$ has nilpotence class 5.

Suppose $G$ has no faithful irreducible character. If $K$ is the kernel of an irreducible character of $G$, then either $c(G/K)\leq 4$, or $c(G/K)=5$ and $G/K$ has maximal class or coclass 2, contradicting Theorem 1.1 (iii) and Lemma 2.10, respectively. Hence we must have $c(G/K)\leq 4$ for all irreducible characters of $G$, which contradicts that $G$ has class 5. Thus $G$ has a faithful irreducible character $\chi$, and by Lemma 2.7, $\chi$ has codegree $p^{a}$.

As $G$ has coclass 3 and nilpotence class 5, we have $|G|=p^{8}$, and by Lemma 2.1, $5\leq a\leq 7$. If $a=7$, then $\chi(1)=p$. By Lemma 2.3 and Theorem 1.1 (i), this is impossible, so $a$ is at most 6. Theorem 1.1 of [3] implies that $\text{cod}(G)=\{1,p,p^{2},p^{a}\}$.

Suppose $p^{2}\leq|Z|\leq p^{3}$. Let $\theta\in\text{Irr}(G/Z)$ have codegree $p^{a}$. Since $a$ is at least 5, $\theta$ is nonlinear, and $p^{a+1}\leq\theta(1)\text{cod}(\theta)\leq|G:Z|\leq p^{6}$. This shows that $a=5$ and $|Z|=p^{2}$. By [5], Theorem 2.47, $G_{4}>Z$ and hence $|G:G_{4}|\leq p^{5}$. Since $c(G/G_{4})=3$, $\text{cod}(G/G_{4})=\{1,p,p^{2},p^{5}\}$, which is impossible. Hence $Z=G_{5}$ has order $p$.

The quotient $G/Z_{2}$ has class 3 and hence $p^{a}\in\text{cod}(G/Z_{2})$. As $|G:Z_{2}|\leq p^{6}$ and $p^{a}=p^{5}$ or $p^{6}$ is the codegree of a nonlinear character of $G/Z_{2}$, we must have $|G:Z_{2}|=p^{6}$ and $a=5$. Let $\gamma\in\text{Irr}(G/Z_{2})$ have codegree $p^{5}$. Then $\gamma$ is nonlinear, so $p^{6}\leq\gamma(1)\text{cod}(()\gamma)\leq|G:Z_{2}|=p^{6}$, which shows that $\gamma$ is a faithful character of $G/Z_{2}$ with degree $p$. By Lemma 2.3, this implies that $\text{cd}(G/Z_{2})=\{1,p\}$.

Theorem 2.47 of [5] implies that $G^{\prime}>Z_{3}$, and by Theorem 1.1 (iii) we may assume $|G:G^{\prime}|\geq p^{3}$. Hence $p^{4}\leq|G^{\prime}|\leq p^{5}$ and $|Z_{3}|\leq p^{4}$. Since $|G:Z_{3}|\geq p^{4}$, Lemma 2.4 implies that $G/Z_{2}$ has an abelian subgroup of index $p$. By Lemma 2.5, we have $p^{6}=|G/Z_{2}|=p|Z_{3}/Z_{2}||G^{\prime}/Z_{2}|$ which shows that $|G^{\prime}|=p^{5}$ and $|Z_{3}|=p^{4}$. As $G/Z_{2}$ has an abelian subgroup of index $p$, $G/Z_{3}$ also has such a subgroup $A/Z_{3}$, and Lemma 12.12 of [6] states that $p^{3}=|A/Z_{3}|=|G^{\prime}/Z_{3}||(A/Z_{3})\cap(Z_{4}/Z_{3})|$. Hence $|(A/Z_{3})\cap(Z_{4}/Z_{3})|=p^{2}$, which shows that $|Z_{4}|=p^{6}$.

Let $x\in G^{\prime}$. Then $z\in Z_{4}$ implies that $[x,g]\in Z_{3}$ for all $g\in G$. Since $|G^{\prime}:Z_{3}|=p$, $x^{p}\in Z_{3}$, and hence passing to $G/Z_{2}$, we have $\overline{1}=\overline{[x^{p},g]}=\overline{[x,g]^{p}}$. This shows that $[x,g]^{p}\in Z_{2}$ for all $x\in G^{\prime}$ and $g\in G$, and we conclude that $G_{3}/Z_{2}$ is elementary abelian. Recall that $G/Z_{2}$ has a faithful character, and hence $Z_{3}/Z_{2}$ is cyclic. Thus $|G_{3}|=p^{3}$.

Let $Z_{3}=\langle a,Z_{2}\rangle$ and observe that $G_{3}=\langle a^{p},Z_{2}\rangle$. Then $[a,g]\in Z_{2}$, and passing to $G/Z$ we have $\overline{[a^{p},g]}=\overline{[a,g]^{p}}=\overline{1}$, which shows that $\overline{a^{p}}$ commutes with all $\overline{g}\in G/Z$, and hence $a^{p}\in Z_{2}$, a contradiction since $G_{3}\neq Z_{2}$. Therefore $G$ must have nilpotence class at most 4. Since $G$ has coclass $3$, we have $n-3=c\leq 4$, and hence $|G|\leq p^{7}$. ∎

For the remainder of this section we will consider only those groups which satisfy Hypothesis $(\ast)$. Together, Lemmas 2.12 - 2.15 prove Theorem 1.3.

Lemma 2.12**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has coclass 4, and $G$ and all of its quotients satisfy Hypothesis $(\ast)$, then the nilpotence class of $G$ is at most 4 and $|G|\leq p^{8}$.

Proof.

Let $G$ satisfy $(\ast)$ and let $|G|$ be minimal such that $G$ has coclass 4 and $|\text{cod}(G)|=4$. Suppose $c(G)=c>5$. The nilpotence class of $G/Z$ is $c-1>4$, so $G/Z$ has coclass at most 3 since $|G|$ was minimal with coclass 4, but this contradicts Theorem 1.1 (iii), Lemma 2.10, and Lemma 2.11. Hence we may assume $G$ has nilpotence class 5.

Suppose $G$ has no faithful irreducible character. If $K$ is the kernel of an irreducible character of $G$, then either $c(G/K)\leq 4$, or $c(G/K)=5$ and $G/K$ has coclass at most 3, contradicting Theorem 1.1 (iii), Lemma 2.10, and Lemma 2.11. Hence we must have $c(G/K)\leq 4$ for all irreducible characters of $G$, which contradicts that $G$ has class 5. Thus $G$ has a faithful irreducible character $\chi$, and by Lemma 2.7, $\chi$ has codegree $p^{a}$. By Lemma 2.1, $a\geq 5$. As $c(G/Z_{2})=3$, we have $p^{a}\in\text{cod}(G/Z_{2})$, and since $G$ satisfies $(\ast)$, the order of $G/Z_{2}$ is at most $p^{6}$, which shows that $a\leq 5$. Hence $a=5$ which forces $Z_{2}$ to have order $p^{3}$.

Let $\gamma\in\text{Irr}(G/Z_{2})$ have codegree $p^{5}$ and notice that $\gamma$ is a faithful character of $G/Z_{2}$ with degree $p$. By Theorem 1.1 (iii), $G^{\prime}$ has order at most $p^{6}$, and $G^{\prime}>Z_{3}$ by [5], Theorem 2.47. Thus Lemmas 2.3 and 2.5 imply that $|G/Z_{2}|=p^{6}=p|G^{\prime}/Z_{2}||Z_{3}/Z_{2}|$. This shows that $|G^{\prime}|=p^{6}$ and $|Z_{3}|=p^{5}$.

Suppose $Z_{2}/Z$ has exponent greater than $p$. Then there exists $x\in Z_{2}$ such that $x^{p}\notin Z$, and $g\in G$ such that $[x^{p},g]\neq 1$. Note that this also implies $|Z|=p$. Since $[x,g]\in Z$, we have $[x^{p},g]=[x,g]^{p}=1$, as $|Z|=p$, a contradiction. Hence $Z_{2}/Z$ is elementary abelian. Repeating this argument in the quotient group $G/Z$, we conclude that $Z_{3}/Z_{2}$ is also elementary abelian, which contradicts that the center of $G/Z_{2}$, $Z_{3}/Z_{2}$, is cyclic with order $p^{2}$. Hence $G$ must have nilpotence class at most 4. Since $G$ has coclass $4$, we have $n-4=c\leq 4$, and hence $|G|\leq p^{8}$. ∎

Lemma 2.13**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has coclass 5, and $G$ and all of its quotients satisfy Hypothesis $(\ast)$, then the nilpotence class of $G$ is at most 4 and $|G|\leq p^{9}$.

Proof.

Let $G$ satisfy $(\ast)$ and let $|G|$ be minimal such that $G$ has coclass 5 and $|\text{cod}(G)|=4$. Suppose $c(G)=c>5$. The nilpotence class of $G/Z$ is $c-1>4$, so $G/Z$ has coclass at most 4 since $|G|$ was minimal with coclass 5, but this contradicts Theorem 1.1 (iii), Lemma 2.10, Lemma 2.11, and Lemma 2.12. Hence we may assume $G$ has nilpotence class 5.

Suppose $G$ has no faithful irreducible character. If $K$ is the kernel of an irreducible character of $G$, then either $c(G/K)\leq 4$, or $c(G/K)=5$ and $G/K$ has coclass at most 4, contradicting Theorem 1.1 (iii), Lemma 2.10, Lemma 2.11, and Lemma 2.12. Hence we must have $c(G/K)\leq 4$ for all irreducible characters of $G$, which contradicts that $G$ has class 5. Thus $G$ has a faithful irreducible character $\chi$, and by Lemma 2.7, $\chi$ has codegree $p^{a}$. Since $G$ has class 5 and coclass 5, we have $|G|=p^{10}$, and by Lemma 2.1, $a\geq 6$. As $c(G/Z_{2})=3$, we have $p^{a}\in\text{cod}(G/Z_{2})$, and since $G$ satisfies $(\ast)$, the order of $G/Z_{2}$ is at most $p^{7}$, which shows that $a\leq 6$. Hence $a=6$ which forces $Z_{2}$ to have order $p^{3}$.

Let $\gamma\in\text{Irr}(G/Z_{2})$ have codegree $p^{6}$ and notice that $\gamma$ is a faithful character of $G/Z_{2}$ with degree $p$. By Theorem 1.1 (iii), $G^{\prime}$ has order at most $p^{7}$, and $G^{\prime}>Z_{3}$ by [5], Theorem 2.47. Thus Lemmas 2.3 and 2.5 imply that $|G/Z_{2}|=p^{7}=p|G^{\prime}/Z_{2}||Z_{3}/Z_{2}|$. This shows that $|G^{\prime}|=p^{7}$ and $|Z_{3}|=p^{5}$.

Suppose $Z_{2}/Z$ has exponent greater than $p$. Then there exists $x\in Z_{2}$ such that $x^{p}\notin Z$, and $g\in G$ such that $[x^{p},g]\neq 1$. Note that this also implies $|Z|=p$. Since $[x,g]\in Z$, we have $[x^{p},g]=[x,g]^{p}=1$, as $|Z|=p$, a contradiction. Hence $Z_{2}/Z$ is elementary abelian. Repeating this argument in the quotient group $G/Z$, we conclude that $Z_{3}/Z_{2}$ is also elementary abelian, which contradicts that the center of $G/Z_{2}$, $Z_{3}/Z_{2}$, is cyclic with order $p^{2}$. Hence $G$ can have nilpotence class at most 4. Since $G$ has coclass $5$, we have $n-5=c\leq 4$, and hence $|G|\leq p^{9}$. ∎

Lemma 2.14**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has coclass 6, and $G$ and all of its quotients satisfy Hypothesis $(\ast)$, then the nilpotence class of $G$ is at most 4 and $|G|\leq p^{10}$.

Proof.

Let $G$ satisfy $(\ast)$ and let $|G|$ be minimal such that $G$ has coclass 6 and $|\text{cod}(G)|=4$. Suppose $c(G)=c>5$. The nilpotence class of $G/Z$ is $c-1>4$, so $G/Z$ has coclass at most 5 since $|G|$ was minimal with coclass 6, but this contradicts Theorem 1.1 (iii) and Lemmas 2.10 - 2.13. Hence we may assume $G$ has nilpotence class 5.

Suppose $G$ has no faithful irreducible character. If $K$ is the kernel of an irreducible character of $G$, then either $c(G/K)\leq 4$, or $c(G/K)=5$ and $G/K$ has coclass at most 5, contradicting Theorem 1.1 (iii) and Lemmas 2.10 - 2.13. Hence we must have $c(G/K)\leq 4$ for all irreducible characters of $G$, which contradicts that $G$ has class 5. Thus $G$ has a faithful irreducible character $\chi$, and by Lemma 2.7, $\chi$ has codegree $p^{a}$. Since $G$ has class 5 and coclass 6, we have $|G|=p^{11}$, and by Lemma 2.1, $a\geq 6$. As $c(G/Z_{2})=3$, we have $p^{a}\in\text{cod}(G/Z_{2})$, and since $G$ satisfies $(\ast)$, the order of $G/Z_{2}$ is at most $p^{8}$, which shows that $a\leq 7$.

Suppose $a=7$. Then $|Z_{2}|=p^{3}$. Let $\gamma\in\text{Irr}(G/Z_{2})$ have codegree $p^{7}$ and notice that $\gamma$ is a faithful character of $G/Z_{2}$ with degree $p$. By Theorem 1.1 (iii), $G^{\prime}$ has order at most $p^{7}$, and $G^{\prime}>Z_{3}$ by [5], Theorem 2.47. Thus Lemmas 2.3 and 2.5 imply that $|G/Z_{2}|=p^{8}=p|G^{\prime}/Z_{2}||Z_{3}/Z_{2}|$. This shows that $Z_{3}/Z_{2}$ has order at least $p^{2}$, and since $\gamma$ is faithful, $Z_{3}/Z_{2}$ is cyclic.

Suppose $Z_{2}/Z$ has exponent greater than $p$. Then there exists $x\in Z_{2}$ such that $x^{p}\notin Z$, and $g\in G$ such that $[x^{p},g]\neq 1$. Note that this also implies $|Z|=p$. Since $[x,g]\in Z$, we have $[x^{p},g]=[x,g]^{p}=1$, as $|Z|=p$, a contradiction. Hence $Z_{2}/Z$ is elementary abelian. Repeating this argument in the quotient group $G/Z$, we conclude that $Z_{3}/Z_{2}$ is also elementary abelian, which contradicts that $Z_{3}/Z_{2}$ is cyclic with order at least $p^{2}$. Hence $a=6$.

Let $\chi\in\text{Irr}(G)$ be faithful with codegree $p^{6}$. Then $\chi(1)=p^{5}$, so $|G:Z|\geq\chi(1)^{2}$ implies $|Z|=p$. As before, this shows that $Z_{2}/Z$ is elementary abelian, and similarly, $Z_{i}/Z_{i-1}$ is elementary abelian for $1\leq i\leq 5$. Since $p^{6}\in\text{cod}(G/Z_{2})$, we have $|G:Z_{2}|\geq p^{7}$, so $|Z_{2}|=p^{3}$ or $p^{4}$. If $|Z_{2}|=p^{4}$, then $G/Z_{2}$ has a faithful character of degree $p$. As in the previous lemmas, $|G:G^{\prime}|\geq p^{3}$ and $G^{\prime}>Z_{3}$. Hence Lemmas 2.3 and 2.5 imply that $|G/Z_{2}|=p^{7}=p|G^{\prime}/Z_{2}||Z_{3}/Z_{2}|$. This shows that $Z_{3}/Z_{2}$ has order at least $p^{2}$, and since $\gamma$ is faithful, $Z_{3}/Z_{2}$ is cyclic. This is impossible since $Z_{3}/Z_{2}$ is elementary abelian. Hence $|Z_{2}|=p^{3}$.

Suppose $|Z_{3}|=p^{4}$. Then $|Z|=p$, for otherwise $G/Z$ contradicts Hypothesis $(\ast)$. Let $N\lhd G$ such that $Z<N<Z_{2}$. Put $Z(G/N)=X/N$ and notice that $Z_{2}\leq X\leq Z_{3}$. If $X=Z_{2}$, then $Z(G/N)=Z_{2}/N$ and $Z_{2}(G/N)=Z_{3}/_{N}$. Since $G/N$ has class 3 or 4, this contradicts Hypothesis $(\ast)$. Hence $X=Z_{3}$. Now $Z_{2}(G/N)=Z_{4}/N$, so $G/N$ has class 3, which implies $N\geq G_{4}$. Since $G_{4}>Z$ by Theorem 2.47 of [5], we have $p^{2}\leq|G_{4}|\leq|N|=p^{2}$, which implies $N=G_{4}$ for all normal subgroups of G lying between $Z$ and $Z_{2}$. This shows that $Z_{2}/Z$ is cyclic, contradicting that it is elementary abelian of order $p^{2}$. Hence $|Z_{3}|\geq p^{5}$.

Now let $H\lhd G$ such that $Z_{2}<H<Z_{3}$ and assume $H\neq G_{3}$. Then $G/N$ has class 3, and $p^{6}\in\text{cod}(G/N)$. Since $|G:N|=p^{7}$, there exists a faithful character of $G/N$ with degree $p$. Put $Y/N=Z(G/N)$ and notice that $Y/N$ is cyclic. The center of $G/N$ lies between $Z_{3}$ and $Z_{4}$, and since $Z_{4}/Z_{3}$ is elementary abelian, this shows that $|Y:N|$ is at most $p^{2}$ and hence $|G:Y|\geq p^{5}$. By Lemmas 2.4 and 2.5 we have $|G/N|=p^{7}=p|G^{\prime}/N||Y/N|\leq p^{3}|G^{\prime}/N|$, which forces $G^{\prime}$ to have order $p^{8}$ and $Y$ to have order $p^{6}$. Since $Y/N$ is cyclic while $Z_{3}/N$ is elementary abelian, $Z_{3}$ must have order $p^{5}$. Lemma 2.4 implies that $G/N$ has an abelian subgroup of index $p$, and hence $G/Z_{3}$ must also have such a subgroup, so we can apply Lemma 2.5 to $G/Z_{3}$. Now $p^{6}=p|G^{\prime}/Z_{3}||Z_{4}/Z_{3}|=p^{4}|Z_{4}/Z_{3}|$, which shows that $Z_{4}$ has order $p^{7}$, which is impossible since $G^{\prime}$ is contained in $Z_{4}$ and has order $p^{8}$. Thus $G$ must have class at most 4. Since $G$ has coclass $6$, we have $n-6=c\leq 4$, and hence $|G|\leq p^{10}$. ∎

Lemma 2.15**.**

Let $G$ be a finite $p$-group with $\text{cod}(G)=\{1,p,p^{b},p^{a}\}$ where $2\leq b<a$. If $G$ has coclass 7, and $G$ and all of its quotients satisfy Hypothesis $(\ast)$, then the nilpotence class of $G$ is at most 4 and $|G|\leq p^{11}$.

Proof.

Let $G$ satisfy $(\ast)$ and let $|G|$ be minimal such that $G$ has coclass 7 and $|\text{cod}(G)|=4$. Similarly to previous lemmas, we may assume $G$ has class 5 and a faithful irreducible character $\chi$ with codegree $p^{a}$. The order of $G$ is $p^{12}$, so $a\geq 7$ by Lemma 2.1. As $G/Z_{2}$ has order at most $p^{9}$, we also have $a\leq 8$.

Suppose first that $a=8$, and hence $|G:Z_{2}|=p^{9}$. Then $G/Z_{2}$ has a faithful irreducible character of degree $p$, and by Lemma 2.3, $\text{cd}(G/Z_{2})=\{1,p\}$. Suppose $Z_{2}/Z$ has exponent greater than $p$. Then there exists $x\in Z_{2}$ such that $x^{p}\notin Z$, and $g\in G$ such that $[x^{p},g]\neq 1$. Note that this also implies $|Z|=p$. Since $[x,g]\in Z$, we have $[x^{p},g]=[x,g]^{p}=1$, as $|Z|=p$, a contradiction. Hence $Z_{2}/Z$ is elementary abelian. Repeating this argument in the quotient group $G/Z$, we conclude that $Z_{3}/Z_{2}$ is also elementary abelian, and since it is also cyclic, we have $|Z_{3}|=p^{4}$. By Lemma 2.5, $p^{9}=p|G^{\prime}/Z_{2}||Z_{3}/Z_{2}|$, and hence $|G^{\prime}|=p^{10}$. This contradicts Theorem 1.1 (iii), so $a=8$ is impossible.

Now $a=7$ and $|G:Z_{2}|=p^{8}$ or $p^{9}$. Suppose $|G:Z_{2}|=p^{8}$. Then $G/Z_{2}$ has a faithful irreducible character of degree $p$, so $\text{cd}(G/Z_{2})={1,p}$ and $Z_{3}/Z_{2}$ is cyclic. The order of $Z_{4}$ is at least $p^{7}$, for otherwise $G/Z_{2}$ would violate Hypothesis $(\ast)$. If $|Z_{3}|=p^{5}$, then by Lemma 2.5, $p^{8}=p|G^{\prime}/Z_{2}||Z_{3}/Z_{2}|$, which implies that $|G^{\prime}|=p^{10}$, contradicting Theorem 1.1 (iii). If $|Z_{3}|=p^{6}$, we have $|G^{\prime}|=p^{9}$. Applying Lemma 2.5 to $G/Z_{3}$, this would imply $|Z_{4}|=p^{8}$, which is impossible since $Z_{4}$ contains $G^{\prime}$. Thus $|Z_{3}|\geq p^{7}$, so $|G/Z_{2}|=p^{8}=p|Z_{3}/Z_{2}||G^{\prime}/Z_{2}|\geq p^{4}|G^{\prime}/Z_{2}|$, which implies that $|G^{\prime}|\leq p^{8}$. Since $G^{\prime}>Z_{3}$ by Theorem 2.47 of [5], this shows that $|G^{\prime}|=p^{8}$ and $|Z_{3}|=p^{7}$. Applying Lemma 2.5 to $G/Z_{3}$, we obtain $|Z_{4}|=p^{10}$.

Notice that $G$ has a faithful irreducible character $\chi$ with codegree $p^{7}$, so $\chi(1)=p^{5}$, and hence $|Z|\leq p^{2}$. Suppose $|Z|=p$. Put $Z_{3}=\langle x,Z_{2}\rangle$, and let $\overline{G}=G/Z$. Since $x^{p^{2}}\notin Z_{2}$, there exists some $g\in G$ such that $\overline{1}\not\equiv\overline{[x^{p^{2}},g]}\equiv\overline{[x,g]^{p^{2}}}\equiv\overline{1}$, since $\overline{[x,g]}\in Z_{2}/Z$, which has order $p^{2}$. Hence $|Z|=p$. Suppose $Z_{2}/Z$ has exponent greater than $p$. Then there exists $y\in Z_{2}$ such that $y^{2}\notin Z$, and $g\in G$ such that $1\neq[y^{p},g]=[y,g]^{p}=1$, a contradiction. Hence $Z_{2}/Z$ is elementary abelian. Repeating this argument in $G/Z$, we see that $Z_{3}/Z_{2}$ is also elementary abelian, which is impossible since $Z_{3}/Z_{2}$ is cyclic of order $p^{3}$. Thus $Z_{2}$ must have order $p^{3}$.

Suppose $Z_{2}/Z$ has exponent greater than $p$. Then there exists $x\in Z_{2}$ such that $x^{p}\notin Z$, and $g\in G$ such that $[x^{p},g]\neq 1$. Note that this also implies $|Z|=p$. Since $[x,g]\in Z$, we have $[x^{p},g]=[x,g]^{p}=1$, as $|Z|=p$, a contradiction. Hence $Z_{2}/Z$ is elementary abelian, and similarly, $Z_{i}/Z_{i-1}$ is elementary abelian for $1\leq i\leq 5$.

Suppose $|Z_{3}|\geq p^{5}$. Let $N\lhd G$ such that $|N|=p^{4}$, $Z_{2}<N<Z_{3}$, and $c(G/N)=3$. Put $X/N=Z(G/N)$, and notice that $Z_{3}\leq X\leq Z_{4}$. Since $G/N$ has class 3, $p^{7}\in\text{cod}(G/N)$, and hence $G/N$ has a faithful irreducible character of degree $p$. Thus $X/N$ is cyclic, and as $Z_{3}/N$ and $Z_{4}/Z_{3}$ are elementary abelian, we have $|X:N|\leq p^{2}$. By Lemmas 2.3 and 2.5, $|G/N|=p^{8}=p|G^{\prime}/N||X/N|\leq p^{3}|G^{\prime}/N|$. Thus $|G^{\prime}|=p^{9}$. If $|Z_{3}|=p^{5}$ or $p^{6}$, then by applying Lemma 2.5 to $G/Z_{3}$ we have $|Z_{4}|=p^{7}$ or $p^{8}$, which is impossible since $Z_{4}$ contains $G^{\prime}$. Thus $|Z_{3}|\geq p^{7}$, which implies $|Z_{2}:Z_{3}|\geq p^{4}$. Since $|X:N|\leq p^{2}$, and $X\geq Z_{3}$, this is impossible. Hence $|Z_{3}|=p^{4}$.

If $|Z|=p^{2}$, then $G/Z$ would violate Hypothesis $(\ast)$, so we have $|Z|=p$. Let $H\lhd G$ such that $Z<H<Z_{2}$. and $c(G/H)=4$. Put $Z(G/H)=Y/H$ and notice that $Z_{2}\leq Y\leq Z_{3}$. If $Y=Z_{3}$, then $G/H$ would have class 3, a contradiction, so $Y=Z_{2}$. Now $Z_{2}(G/H)=Z_{3}/H$, which contradicts Hypothesis $(\ast)$. Hence $G$ must have class at most 4. Since $G$ has coclass $7$, we have $n-7=c\leq 4$, and hence $|G|\leq p^{11}$. ∎