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Are there any examples of groups, $G$, such that $|G|=p^aq^b$ (where $p$ and $q$ are distinct primes and $a,b\geq 1$) and $|\mathrm{Z}(G)|=p^a$?

($\mathrm{Z}(G)$ denotes the center of $G$)

I tried seeing if such a group could be constructed by a more general method - by taking the direct product of two groups, $X$ and $Y$, where $|X|=p^a$, $|Y|=q^b$, and $X$ is abelian.

But, of course, $\mathrm{Z}(X\times Y)=\mathrm{Z}(X)\times\mathrm{Z}(Y)$, and since $|Y|$ is a prime power, $Y$ must have a non-trivial center, meaning that $|\mathrm{Z}(X\times Y)|$ will always be some multiple of $p^a$ rather than just $p^a$.

I also looked at some small groups, but I couldn't find any example (because most were Dihedral or Dicyclic groups with trivial/ near-trivial centers).

(I am aware of Burnside theorem, but I can't figure out anything substantive from it in this situation)

CardioidAss22
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  • @ArsenBerk Sorry, I'll edit the question. – CardioidAss22 Jun 08 '19 at 17:51
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    This is not possible. The Sylow $p$-subgroup being central implies that it is a direct factor and hence the sylow $q$-subgroup also contributes to the center. – Tobias Kildetoft Jun 08 '19 at 17:58
  • @TobiasKildetoft Can you please elaborate? Are you saying that $G$ will be the direct product of its $p$ and $q$-Sylow subgroups? If so, how can we show that the $q$-Sylow subgroup is a normal one? – CardioidAss22 Jun 08 '19 at 18:06
  • The Sylow $q$-subgroup is normalized by $Q$ and by $P$ and hence by all of $G$. – Derek Holt Jun 08 '19 at 20:32
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    Here is a more detailed version of what Tobias is saying. Let $X$ be a Sylow $p$-subgroup and $Y$ a Sylow $q$-subgroup. Now $G=XY$ and by your hypothesis $X\leq Z(G)$, so $Z(Y)\leq Z(G)$. On the other hand $Z(Y)\neq 1$ (this is a general property of groups of prime power order) so $q$ divides $|Z(G)|$. – verret Jun 08 '19 at 22:53
  • An easier follow-up exercise. Give an example of a finite non-abelian group such that its center is also one of its Sylow subgroups. – Jyrki Lahtonen Apr 04 '22 at 06:39

2 Answers2

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Attempt by using the guide given by Tobias Kildetoft:

Suppose that $|Z(G)|=p^a$.
Note that $Z(G)$ is a Sylow $p$-subgroup of $G$, hence I denote it by $P$.

By Sylow First Theorem, there is a Sylow $q$-subgroup $Q$.
By Product Formula, $G=PQ$. And clearly $P\cap Q=1$.
And since $P=Z(G)$, $P\le N_G(Q)$. Hence $G=N_G(Q)$ which claims that $Q\lhd G$.
These show that $$G=P\times Q$$ Thus $$Z(G)=Z(P)\times Z(Q)$$ Since $Q$ is a $q$-group, $Z(Q)\neq 1$. Hence $|Z(G)|\ne p^a$, a contradiction.

Wang Kah Lun
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Unless this is a Sylow/direct products exercise, there's a way more elementary argument. If $|Z(G)|=p^a$, then every noncentral element has centralizer of order $p^aq^{b'}$, for some $0< b'<b$; therefore, the class equation yields $p^aq^b=p^a+\sum_iq^{b-b'_i}$: contradiction, because $q$ divides the LHS and $\sum_i$ at the RHS, but not $p^a$.

Kan't
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