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No one of the 38 nonabelian groups of order $112$ (up to isomorphism) is centerless. Indeed, this answer uses advanced tecniques (of course w.r.t. my level of knowledge, such as solvability, Burnside's $p^aq^b$-theorem, etc.) to show a stronger result, namely that every group of order $112$ has an element of order $2$ in its center.

Is there any "lower-tech" argument to prove the (weaker than the above mentioned one) result in the title? For example, I expected that the class equation in the centerless case to lead to some contradiction, also considering the possibility that any between the $2$-Sylow or the $7$-Sylow was normal, but unsuccessfully as far as I could go (both $16=1+2^3+7$ and $7=1+2+2+2$ are allowed sizes of union of conjuagacy classes). The only I could get to, is that the number of conjugacy classes of size $7$ must be odd, but that's seemingly not enough to get any contradiction.

Kan't
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  • Perhaps I'm missing anything, but doesn't that argument just prove that the center cannot have order $16$, @DietrichBurde? – Kan't Aug 21 '23 at 11:55
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    You are right, $p^a=2^4$ is $16$ in the example given there, with $Y=C_7$ abelian. But the proof gives possibly more. Certainly we need some assumption on the order $p^aq^b$ of $G$, since, say, for $p=q=1$ the group may have indeed trivial center. – Dietrich Burde Aug 21 '23 at 12:27
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    Here orders $n$ are studied, where there exists a group of order $n$ with trivial center - so the complement. – Dietrich Burde Aug 21 '23 at 13:14

2 Answers2

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Let $G$ be a group of order $p^aq^b$, with different primes $p$ and $q$. The general question is, which conditions on $p,q,a,b$ are needed, so that every group of of order $p^aq^b$ has a non-trivial center. Clearly there must exist some conditions, since not every group of, say, order $pq$ has a non-trivial center.

Certainly, if $G$ is nilpotent, the center is always non-trivial. So we may assume that $p^aq^b$ is not a "nilpotent number".

In fact, OEIS A216594 gives a table of non-nilpotent positive integers $n$ such that every group of order $n$ has a non-trivial center. The sequence of such $n$ starts with $$ 28, 40, 44, 63, 76, 88, 92, 104, 105, 112, 117, 124, 152,\cdots $$ Also $n=112$ is contained in the list. There seems not too much known for the general case, except the following. If $p$ is any prime with $p\equiv 3\bmod 4$, then $n=4p$ is in the list.

This post gives some conditions for a non-trivial center for groups of order $p^aq^b$, using Sylow subgroups. However, it seems that this only concerns the case $a\le 2$ or $b\le 2$, where every group of order $p^2$ or $q^2$ is abelian.

So for $112=2^4\cdot 7$, this post is relevant, even if the methods may not be so elementary. You don't need Burnside to show that every group of order $112$ is solvable - see here:

Prove that group of order 112 is solvable

Dietrich Burde
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  • Thank you for your input. Yes, the "general question" is exactly as you have posed here, though, in view of my question, we can safely assume $b=1$: it's precisely this latter specialization, which I was hoping to bring the matter away from (relatively) hi-tech realms as solvability, Burnside's $p^aq^b$-theorem, etc. But probably it's just time for me to finally look at them, at least to solvable groups... – Kan't Aug 21 '23 at 13:22
  • Solvability is not hi-tech. It is very basic group theory. But of course, Burnside's theorem would be overkill, at least for order $112$. Did you see the article on "Trivial Center (TZ) Groups"? – Dietrich Burde Aug 21 '23 at 13:24
  • Not yet, going to. – Kan't Aug 21 '23 at 13:25
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To clarify, the answer linked in the question shows that every group $G$ of order $|G| = 112$ has an element of order $14$.

The fact that $Z(G)$ always has an element of order $2$ follows with a few more arguments. Here is a proof:

If a $7$-Sylow $R$ is not normal, the existence of an element order $14$ implies that $N_G(R)$ is cyclic of order $14$. By transfer (Burnside's transfer theorem) there exists a normal complement for $R$.

In other words, there is a normal Sylow $2$-subgroup $P \triangleleft G$. Here $Z(P)$ has order $2$, $4$, or $16$, so $R$ centralizes a nontrivial element of $Z(P)$; such an element is contained in $Z(G)$.

If $R$ is normal, then $C_G(R) = Q \times R$ where $Q$ is a nontrivial normal $2$-subgroup of $G$. Take a Sylow $2$-subgroup $P$ containing $Q$; then $Q \cap Z(P)$ is nontrivial and contained in $Z(G)$.

Mikko Korhonen
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