A capable group is a group that is isomorphic to $G/Z(G)$ for some group $G$.

So, is every member of the OEIS sequence A216594 (consisting of those $n$ for which there is a non-nilpotent group, but no centerless group, of order $n$) the order of a capable group?

Every even number $2n$ (for $n>1$) is the order of a capable group. Indeed, the dihedral group of order $2n$ is isomorphic to the central quotient of the dihedral group of order $4n$ as pointed out at Non-cyclic numbers that are not the order of any capable group.

So, only odd numbers need to be considered. The first odd member of A216594 is $63$. Since $63$ is not a powerful number (it is divisible by the prime $7$ but not by $7^2=49$), there is no capable abelian group of order $63$ by the known characterization of finite capable abelian groups (isomorphic to a direct sum of cyclic groups, with the order of each one dividing the order of the next one and the last two being isomorphic, or equivalently, having the same order).

So, to try to find a capable group of order $63$, one must consider nonabelian groups. If $G$ is the nonabelian group of order $21$ (given as the semidirect product $\mathbb{Z}_7 \rtimes \mathbb{Z}_3$ with $(a, b)(c, d)=(a+2^bc, b+d)$), then one could consider the direct product $G \times \mathbb{Z}_3$. The other nonabelian group of order $63$ is given by the semidirect product $\mathbb{Z}_7 \rtimes \mathbb{Z}_9$ (again with $(a, b)(c, d)=(a+2^bc, b+d)$). If neither of those two nonabelian groups of order $63$ turns out to be capable, then the answer to the question must be "no".

Otherwise, one could consider the next odd member of A216594, namely $105$, which is also the first squarefree member. The only nonabelian group of order $105$ is given by the direct product of the nonabelian group of order $21$ with the cyclic group of order $5$. If that group of order $105$ turns out to not be capable, then the answer to the question must again be "no".

If neither $63$ nor $105$ succeeds in finding a member of A216594 that is not the order of any capable group, then one could consider $117$ (the next odd member) or $165$ (the next odd squarefree member). But $117$ and $165$ have the same "nilpotent number relationships" (of the form $p^k \equiv 1 \pmod q$) among their prime factors and the same prime signature as $63$ and $105$ respectively, so I do not think that $117$ or $165$ would succeed if $63$ and $105$ did not succeed. So, the next order to consider would be $189$.

  • $\begingroup$ Any nilpotent group (not just abelian groups) of order a non-powerful number cannot by capable, as a finite nilpotent group is capable if and only if its $p$-parts are. $\endgroup$ Mar 4 at 4:04

2 Answers 2


The group of order $63$ with elementary abelian Sylow $3$-subgroup is capable - it is isomorphic to $G/Z(G)$ with $|Z(G)|=3$.

However the nonabelian group of of order $105$ is not capable, because it has the form $H \times C_p$ where $|H|$ is coprime to the prime $p$.

Suppose such a group was isomorphic to $G/Z(G)$. Then the inverse image of the $C_p$ factor of $G/Z(G)$ would be abelian and normal in $G$, and $G$ would have a normal Sylow $p$-subgroup $P$. A complement of $P$ is $G$ would centralize both $P \cap Z(G)$ and $P/P \cap Z(G)$ and hence centralize $P$, so $P$ would be a direct factor of $G$ and hence in $Z(G)$, contradicting the structure of $G/Z(G)$.


To generalize Derek Holt's observation, I claim that any group of the form $G=H\times C_q$ with $q\gt 1$ and $\gcd(|H|,q)=1$ cannot be capable. I believe this follows from a theorem of Beyl, Felgner and Schmidt ("On groups occurring as center factor groups", J. Algebra, 61 (1979)) which gives sufficient conditions under which the capability of a direct product of finitely many groups would imply the capability of the direct factors, but below I give a direct argument. Note that $q$ need not be squarefree, so this is a more general result.

Indeed, let $K$ be a group and $N\leq Z(K)$ be a central subgroup such that $K/N\cong G= H\times C_q$. It is known that if a finite group is capable then it is the central quotient of a finite group, so we may take $K$ to be finite. We will show that $N\neq Z(K)$.

Let $M$ be the preimage of $C_q$ in $K$; then $M/N$ is cyclic, with $N$ central, so $M$ is abelian. Since the image is normal in $G$, then $M$ is normal in $K$. Moreover, if $y\in K$ has order prime to $|H|$ (in particular, if it has order that divides a power of $q$), then its image lies in the $C_q$ factor, so $y\in M$.

Let $x$ be an element that maps to a generator of the $C_q$ factor of $G$. We may assume the order of $x$ divides a power of $q$: if the order is $rs$ with $\gcd(r,s)=\gcd(r,q)=1$, then $x^r$ has order $s$ (which divides a power of $q$); it maps to the $r$th power of the original generator, and since $\gcd(r,q)=1$, this element also generates.

I claim that $x$ commutes with every element whose order is relatively prime to $q$. Indeed, if $h$ has order $r$, $\gcd(r,p)=1$, then $1=[hN,xN]=[h,x]N$, so $[h,x]\in N\leq Z(K)$. In particular, $[h,x]\in Z(\langle h,x\rangle)$, so $\langle h,x\rangle$ is nilpotent of class at most $2$. Since $h$ and $x$ have relatively prime order, they commute.

We also know that $x$ commutes with every element of order dividing a power of $q$, since they all lie in $M$ and $M$ is abelian.

Now let $k\in K$ be arbitrary, and let its order be $r$. Write $r=st$ with $\gcd(s,t)=\gcd(s,q)=1$. Let $u,v\in\mathbb{Z}$ be such that $1=us+vt$. Then $$ k=k^{us+vt} = (k^s)^{u}(k^{t})^v.$$ Note that $k^s$ has order $t$, so $k^{su}$ has order that divides a power of $q$. And $k^{t}$ has order $s$, so $k^{vt}$ has order prime to $q$. Thus $k$ is a product of elements that centralize $x$, and hence $k$ itself centralizes $x$.

Thus, $x\in Z(K)$. However, $xN$ is nontrivial, so $N\neq Z(K)$, which is what we needed to prove.

Note that if $n$ is not a nilpotent number, then neither is any positive multiple of $n$. This gives a way of constructing non-nilpotent numbers that are not the order of any capable group: take any non-nilpotent number $m$, and find a prime $p$ such that:

  1. $\gcd(m,p)=1$.
  2. If $d$ is a positive integer such that $d\mid m$ and $d\equiv 1\pmod{p}$, then $d=1$.
  3. $\gcd(m,p-1)=1$.

Let $n=mp$. If $G$ is a group of order $n$, then it has the form $H\times C_p$. Indeed, conditions 1 and 2 ensure the Sylow $p$-subgroup of $G$ is normal of order $p$, and condition 3 ensures the action of any element of $G$ on this subgroup by conjugation is trivial. Since the Sylow $p$-subgroup of $G$ is central, then by Burnside's Normal $p$-complement Theorem, the subgroup has a normal $p$-complement $H$ in $G$. And since the Sylow $p$-subgroup is normal, we conclude that $G=H\times C_p$ (Burnside's theorem says that if a Sylow $p$-subgroup of $G$ lies in the center of its normalizer, then it has a $p$-complement).

We can find non nilpotent numbers of this type by taking any odd non-nilpotent number $m$, and picking a prime of the form $p=km+2$. If $p\gt m$ we get properties 1 and 2. To see $3$, let $q$ be a prime dividing $m$. Then $p-1=km+1\equiv 1\pmod q$, so $q\nmid p-1$. So $\gcd(m,q-1) =1$. Then $n=mp$ is not a nilpotent number.

And of course there are other orders in which we are guaranteed that a Sylow subgroup is of prime order and central in any group of that order, even if it is not always for the same prime, and even if it is not an odd nilpotent number. For $n=105$, if there is a normal Sylow $5$-subgroup then it is central; if it is not normal, then there are $21$ of them, then that accounts for $80$ nontrivial elements, which means the Sylow $7$-subgroup is normal and central (as if there is more than one, there are $15$ of them...)

  • $\begingroup$ I don't understand the first sentence - doesn't Derek Holt's answer prove that $H \times C_p$ for $\gcd(|H|,p) = 1$ is not capable? $\endgroup$ Mar 4 at 15:00
  • $\begingroup$ @testaccount I meant to change that to any cyclic factor of order prime to $|H|$,not just prime order. $\endgroup$ Mar 4 at 15:57
  • $\begingroup$ @testaccount fixed. $\endgroup$ Mar 4 at 16:15
  • $\begingroup$ I see, thanks. (With the same notation, and especially $p$ suggests a prime..) $\endgroup$ Mar 5 at 1:41
  • $\begingroup$ @testaccount I originally thought, "any cyclic factor prime to $|H|$"; then I chickened out but didn't correct it properly, then I realized, yes any cyclic factor and didn't correct it properly and used bad notation. I think it's fixed now. $\endgroup$ Mar 5 at 2:19

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