If $G=Z(G)$, then the statement is true. So suppose that $G\neq Z(G)$.
Then by the class equation, we have
\[|G|=|Z(G)|+ \sum_{i=1}^r |G:C_G(g_i)|,\]
where $g_i$ are representatives of the distinct conjugacy class not contained in the center $Z(G)$, and $C_G(g_i)$ is the centralizer of $g_i$.
(Since we are assuming that $G \neq Z(G)$ such $g_i$ exist.)

Since $g_i \not \in Z(G)$, the groups $C_G(g_i)$ are proper subgroups of $G$ and hence $p$ divides $|G: C_G(g_i)|$. Of course $p$ divides $|G|$, thus $p$ should divide $|Z(G)|$ as well.
Therefore the center $Z(G)$ cannot be trivial.

Comment.

This problems is a simple/nice application of the class equation of group theory. The only information about the group is that its order is a prime power. From this we can conclude that the center of the group is not trivial.

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Proof.
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Hint.
You may directly compute the conjugates of each element
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Hint.
It follows from […]

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Using the generators and relations, we have
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\]
(a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$.
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