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Let $G$ be a group such that $\# G - \# \text Z(G) = 46$. The problem is to figure out $\text{ord}(G)$. As $\text Z(G) \leq G$, Lagrange's Theorem tells us that $\# \text Z(G) \mid \text{ord}(G)$. Therefore, $\exists k \in \mathbb N, k > 1$ such that $\text{ord}(G) = k \cdot \# \text Z(G)$

This gives us $$k \cdot \# \text Z(G) - \# \text Z(G) = 46$$ or in other words $$\# \text Z(G) = \frac{46}{k - 1}$$ As this must be a natural number, there exist a priori only 4 values $\# \text Z(G)$ can possibly take: $\# \text Z(G) \in \{1, 2, 23, 46\}$.

We can exclude $\# \text Z(G) = 1$, as this would lead to $\text{ord}(G) = 47$, a prime number. But the center of a $p$-group cannot be trivial, so $1$ is off the list.

I suspect there are similar criteria to eliminate two of the remaining three values, but I can't think of any at the moment. Could someone maybe tell me a theorem to solve this?

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    $\begingroup$ You could also look at $G/Z(G)$ and count the possibilities there as well. For example, if $\# Z(G)=46$, then this quotient is cyclic of order $2$. But this implies $G$ is abelian, contradiction. This also applies to the case the center is order $23$. $\endgroup$
    – Shrugs
    Nov 22, 2023 at 0:03
  • $\begingroup$ Too late to add to the comment, but the dihedral group of order $48$ gives the case of center being order $2$. $\endgroup$
    – Shrugs
    Nov 22, 2023 at 0:08
  • $\begingroup$ @K02 Of course, thank you! I really should've had that idea on my own... The example is also really nice! $\endgroup$
    – Minerva
    Nov 22, 2023 at 0:37
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    $\begingroup$ Why use two notations $\#G$ and ord$(G)$ for the same quantity? $\endgroup$ Nov 22, 2023 at 3:04
  • $\begingroup$ @GregMartin I used the notation as it was used in the question, although I agree that it is unnecessary. $\endgroup$
    – Minerva
    Nov 22, 2023 at 15:05

1 Answer 1

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Since my comments seem to more or less answer OP's question, let me write out a detailed answer so that this question doesn't join the list of unanswered questions.

Suppose $\#G-\#Z(G)=46$. Then as OP observed, $\#Z(G)=\frac{46}{k-1}$ where $k:=\frac{\#G}{\#Z(G)}$. In particular, $k$ is the order of the quotient group $G/Z(G)$. It is known that if $G/Z(G)$ is cyclic, then $G$ itself must have been abelian and so $Z(G)=G$.

Since $k-1$ must divide $46$, and the divisors of $46$ are $1,2,23,46$, we can are tasked with ruling out as many cases as possible. For $k-1=46$, we find that $\#G=47$ and which is a prime. So $G$ is cyclic of prime order and therefore $Z(G)=G$ which is absurd. For $\#Z(G)=23$, we know from $\#G-\#Z(G)=46$ that $\#G/Z(G)$ is of order $k=3$. Ang group of order $3$ is cyclic so $\#Z(G)=\#G>\#Z(G)$ which is absurd. Similarly, if $\#Z(G)=46$ then $k=2$ and this is absurd for the same reason.

The only permissible value of $k$ is $k=24$ which occurs when $Z(G)$ is cyclic of order $2$. In this case, $\#G-\#Z(G)=46$ implies that $G$ must be an order $48$ group with $Z(G)\cong \mathbb{Z}/2\mathbb{Z}$. Such groups exists and that is the dihedral group $D_{24}$ of order $48$ (and another is the simpler example @kabenyuk gives of $S_4\times \mathbb{Z}/2\mathbb{Z}$).

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  • $\begingroup$ Good answer. There are $15$ groups of order $48$ with center of two elements. In my opinion the simplest example is the group $S_4\times\mathbb{Z}_2$. $\endgroup$
    – kabenyuk
    Nov 22, 2023 at 7:49

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