# $\# G - \# \text Z(G) = 46$. What is the value of $\text{ord}(G)$?

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?

• 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 $23$. Nov 22, 2023 at 0:03
• Too late to add to the comment, but the dihedral group of order $48$ gives the case of center being order $2$. Nov 22, 2023 at 0:08
• @K02 Of course, thank you! I really should've had that idea on my own... The example is also really nice! Nov 22, 2023 at 0:37
• Why use two notations $\#G$ and ord$(G)$ for the same quantity? Nov 22, 2023 at 3:04
• @GregMartin I used the notation as it was used in the question, although I agree that it is unnecessary. Nov 22, 2023 at 15:05

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 $$S_4\times \mathbb{Z}/2\mathbb{Z}$$).
• Good answer. There are $15$ groups of order $48$ with center of two elements. In my opinion the simplest example is the group $S_4\times\mathbb{Z}_2$. Nov 22, 2023 at 7:49