For a non-commutative group, can the center have a non-trivial subgroup?

Question

Let $G$ be a non-commutative group, and $Z$ be the center of $G$. Can $Z$ have a non-trivial subgroup? I can't prove the negative, but I can't find an example either.

A line of thought I'm currently pursuing is that $Z$ is still a group, so one of Sylow's theorem (obviously we need that $G$ is a finite group here, but this is enough as an example) says that there must be some $p$-groups, and we have the example. However I do not know a specific example where the center is not of order $p^m$ (order $p_1^{n_1}p_2^{n_2}$ is good).

Another proof is preferable but an example of a group with center of order $p_1^{n_1}p_2^{n_2}$ is also an interesting answer.

Answer 1

Try $\Bbb Z / 6 \Bbb Z \times S_3$.

Answer 2

The smallest nonabelian group is $S_3$ up to isomorphism. The centre of $S_3$ is trivial. In a direct product $H=A\times S_3$ of an abelian group $A$ with $S_3$, the centre is $\{(a,e_{S_3})\mid a\in A\}$ (which is isomorphic to $A$), since for any $(x,y)\in H$, we have

$$\begin{align} (x,y)(a,e_{S_3})&=(xa,ye_{S_3})\\ &=(ax,e_{S_3}y)\\ &=(a,e_{S_3})(x,y). \end{align}$$

Therefore, it suffices to consider the smallest abelian $A$ for which $A$ has a nontrivial subgroup. There are thus two choices using this method:

$$\Bbb Z_4\times S_3\text{ or }(\Bbb Z_2)^2\times S_3.$$

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My initial answer:

The group given by

$$G = \langle a,b \mid a^8,b^2,bab=a^5 \rangle$$

has centre isomorphic to $\Bbb Z_4$, which, in turn, has a subgroup isomorphic to $\Bbb Z_2$. See here.

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PS: The group $G$ above has order $16$, which is less than $4\times (3!)=24$. Therefore, the technique described above is not exhaustive and does not give the smallest possible example.