A normal subgroup of $G$ is a subgroup of the center of $G$. I do not know how to answer question 10.31 from Dan Saracino's Abstract Algebra: A First Course. The question is the following,
Suppose that $p$ is prime, $n$ is a positive integer, and $G$ is a group of order $p^n$. Prove that if $H$ is a subgroup of order $p$ and $ghg^{-1}$ is in $H$ for all $g$ in $G$ and all $h$ in $H$, then $H$ is a subgroup of $Z(G)$ (i.e. the center of $G$).
I should mention that up to this point, we have not learned about normal subgroups at all. I just saw the definition on the next page and noticed that H is a normal subgroup. In this chapter, we have only learned about Lagrange's Theorem and the Class Equation.
Here is my attempt at a solution.
For fixed $g$ in $G$, $gHg^{-1}$ is a subgroup of $H$. But, by Lagrange's theorem, this means that $gHg^{-1}$ has order $p$. This means $gHg^{-1} = H$. That is, $gH = Hg$. Also, $H$ is a cyclic group. This means $gh^{k} = h^{m}g$. I really do not know how to proceed. I do not see how we can use the class equation here either really.
Any help with how to proceed would be great.
 A: As, $H$ is normal in $G$, hence, for any $g\in G$, $ghg^{-1}\in H \implies ghg^{-1}=h_1 $ , for some $h,h_1\in H $
Now we go for the conjugacy class for each element $h'\in H $.
Now, as $|H|=p$, so, conjugacy class of any element $h'$ in $H$ can contain at most $(p-1)$ elements.
But , we should remember that order of each class must divide $|G|=p^{n}$.
So,  each conjugacy class of each element in $H$ contains exactly one element, as $1$ is the only member in $1,2,\cdot\cdot\cdot,(p-1)$, which divides $p^{n}$.
So, that means for any $g\in G $, $ghg^{-1}=h$ for any $h\in H $.
So, $H⊂Z(G)$
A: If $H\unlhd G$, then we can consider the action of $G$ on $H$ by conjugation, which leads to the following "orbit equation":
$$|H|=|H \cap Z(G)|+\sum_{h \in \{Orbits \space rep's\}}\frac{|G|}{|C_G(h)|} \tag 1$$
where "$Orbits$" (capital "O") stands for the orbits of size greater than $1$, if any. In that case, all the terms in the "$\sum$" in $(1)$ would be of the form $p^{\alpha}$, with $\alpha>1$ (because, by the Orbit-Stabilizer Theorem, $|O(h)|>1\Rightarrow |C_G(h)|<|G|$). Now, by assumption $|H|=p$, thence $|H\cap Z(G)|$ is either $1$ or $p$; the former case is ruled out by $(1)$ and the subsequent discussion, so we are left with $|H\cap Z(G)|=|H|$ (and no nonunit orbits), whence $H\le Z(G)$.
A: Every $p-$group is nilpotent and $H\lhd G$ so $H\cap Z(G)\not=1$ and since $H$ is of prime order it has to be $H\cap Z(G)=H$
