# Non-cyclic numbers that are not the order of any capable group

A group $$G$$ is said to be capable if there is some group $$H$$ for which $$H/Z(H)$$ is isomorphic to $$G$$.

It is known that the only capable cyclic group is the trivial group. So, if $$n$$ is a cyclic number (i.e., the cyclic group is the only group of order $$n$$) other than $$1$$, then there is no group whose center has index $$n$$.

But is there a non-cyclic number $$n$$ for which there is no group whose center has index $$n$$ (or equivalently, there is no capable group of order $$n$$)?

Checking small values of $$n$$:

• $$n=4$$: The Klein four-group is capable (arising as the central quotient of the quaternion group).
• $$n=6$$: The symmetric group $$S_3$$ is capable (arising as its own central quotient, as it is centerless).
• $$n=8$$: A finite abelian group is known to be capable if and only if it is isomorphic to a finite direct sum $$\mathbb{Z}_{n_1} \oplus \mathbb{Z}_{n_2} \oplus ... \oplus \mathbb{Z}_{n_k}$$ where $$n_1 \mid n_2 \mid ... \mid n_{k-1} = n_k$$ (i.e., the last two invariant factors coincide). So, the elementary abelian $$2$$-group of order $$8$$ is capable.
• $$n=9$$: The elementary abelian $$3$$-group of order $$9$$ is capable by the same argument.
• $$n=10$$: The dihedral group of order $$10$$ is centerless, so it is capable.
• $$n=12$$: The alternating group $$A_4$$ is centerless, so it is capable.
• $$n=14$$: The dihedral group of order $$14$$ is centerless, so it is capable.
• $$n=16$$: The elementary abelian $$2$$-group of order $$16$$ is capable by the same argument as for orders $$8$$ and $$9$$.
• $$n=18$$: The dihedral group of order $$18$$ is centerless, so it is capable.
• $$n=20$$: There is a centerless group of order $$20$$, which is then also capable.

So far, no such $$n$$ has been found yet.

To find numbers that are not the order of centerless groups, look up A216594 and A056867 in the OEIS. The smallest number in A056867 that is neither a cyclic number nor a power of a prime is $$45$$, while the smallest number in A216594 is $$28$$.

This means that the smallest $$n$$ that answers the question (if it exists) could be $$28$$ or $$45$$ or another member of A216594 or A056867.

• If I'm not mistaken, $D_{56}$, the dihedral group, rules out $28.$ Feb 27 at 6:18
• All the $n!$'s are ruled out. $S_n$ is centerless, $n\ge3.$ Feb 27 at 6:29
• Dihedral groups wipe out all evens. Feb 27 at 6:48

A number $$n$$ with prime factorization $$n=p_1^{a_1}\cdots p_r^{a_r}$$ where the $$p_i$$ are distinct primes and $$a_i\geq 1$$ for all $$i$$, is a nilpotent number if and only if for all $$1\leq i,j\leq r$$, and $$1\leq k\leq a_i$$, $$p_i^k\not\equiv 1\pmod{p_j}$$.

A theorem of Pazderski from 1959 shows that every group of order $$n$$ is nilpotent if and only if $$n$$ is a nilpotent number. It is a theorem of Dickson from 1905 that every group of order $$n$$ is abelian if and only if $$n$$ is a cube free nilpotent number. It is a theorem of Szele from 1947 that every group of order $$n$$ is cyclic if and only if it is a squarefree nilpotent number. (See this post)

It is a corollary to a Theorem of Baer that a finite abelian group is capable if and only if it is not cyclic, and its two largest invariant factors are equal. That is, if $$A$$ is a finite abelian group written in the form $$A= C_{m_1}\oplus\cdots\oplus C_{m_r},\qquad m_1\mid m_2\mid\cdots\mid m_r,$$ with $$C_k$$ the cyclic group of order $$k$$, the $$A$$ is capable if and only if $$r\gt 1$$ and $$m_{r-1}=m_r$$.

So take $$n=45 =3^2\times 5$$. By Szele's theorem, not every group of order $$45$$ is cyclic. By Dickson's theorem every group of order $$45$$ is abelian. And the two abelian groups of order $$45$$ are $$C_{45}$$ and $$C_3\oplus C_{15}$$, with neither capable by Baer's characterization of capable groups that are direct sums of cyclic groups.

In fact, $$45$$ is the smallest non-cyclic number in which all groups are not capable.

For $$k\geq 2$$, the dihedral group of order $$4k$$ has center cyclic of order $$2$$, and the central quotient is the dihedral group of order $$2k$$. So every even number greater than $$2$$ is the order of a capable group, namely the dihedral group of that order.

For prime powers $$p^k$$, $$k\geq 2$$, we can always take the elementary abelian $$p$$-group of that order, which is capable.

For orders of the form $$pq$$ with $$p$$ and $$q$$ primes, and which are not cyclic numbers, the nonabelian group of order $$pq$$ has trivial center, so it is equal to its own central quotient.

For orders less than $$45$$:

1. For even number $$2k$$, $$2\leq k\leq 22$$, there is a capable group of order $$2k$$.

2. The values $$2$$, $$3$$, $$5$$, $$7$$, $$11$$, $$13$$, $$15$$, $$17$$, $$19$$, $$23$$, $$29$$, $$31$$, $$33$$, $$35$$, $$37$$, $$41$$, and $$43$$ are cyclic numbers.

3. The values $$9$$, $$25$$, and $$27$$ are prime powers that are not prime, so there are capable groups of that order.

4. The nonabelian groups of order $$21$$ and $$39$$ are centerless and capable.

So $$45$$ is the smallest number that is not a cyclic number, but in which every group of order $$45$$ is not capable.

It it not hard to show that a finite nilpotent group is capable if and only if its $$p$$-parts are capable. Thus, for nilpotent numbers $$n$$, every group of that order is not capable if and only if there is a prime $$p$$ such that $$p\mid n$$ but $$p^2\nmid n$$.

And any number of the form $$p_1^{a_1}\cdots p_r^{a_r}$$, with $$p_i$$ distinct primes and $$a_j\geq 2$$ for all $$j$$ is the order of a capable group (just take a direct product of the corresponding elementary abelian groups).

I would think that non-nilpotent numbers are always the order of a capable groups, but I would not necessarily be surprised if it turns out that this is not the case. I haven't really thought it through.

• Are groups of order a nilpotent number nilpotent? Feb 27 at 7:26
• @metoo As it says in the linked-to post, every group of order $n$ is nilpotent if and only if $n$ is a nilpotent number. That's a theorem of Pazderski (1959) Feb 27 at 7:28
• For the conjecture in the last paragraph, see math.stackexchange.com/questions/4873980/…. Mar 2 at 16:10
• Unfortunately, not every non-nilpotent number is the order of a capable group, as the answer to the question linked in the above comment shows. For example, the non-nilpotent number $105$ is not the order of any capable group. Mar 2 at 21:05
• @GeoffreyTrang Calling it a "conjecture" gives me too much credit. It was little more than a first impression. I even said I wouldn't be surprised if it wasn't the case. And I don't need up-to-the-minute updates. Mar 2 at 21:11