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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.

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    $\begingroup$ I believe there's a typo here: "Prove that if H is a subgroup ... then H is a subgroup of G." $\endgroup$
    – lisyarus
    Sep 8 '20 at 14:46
  • $\begingroup$ @John Wick The timing I'm afraid is not at all proper to be tackling such problems if you haven't even yet studied normal subgroups. $\endgroup$
    – ΑΘΩ
    Sep 8 '20 at 14:53
  • $\begingroup$ Do you know about group actions? $G$ acts (via conjugation) on $H$. What does $\lvert G\rvert = p^n$ tell you about the size of the orbits? (If you don't know about group actions yet, use the class equation on $G$, what do you know about the size of the conjugacy classes of elements of $H$?) $\endgroup$ Sep 8 '20 at 14:58
  • $\begingroup$ @DanielFischer If OP hasn't learnt about normal subgroups, they probably haven't learnt group actions. $\endgroup$
    – player3236
    Sep 8 '20 at 14:59
  • $\begingroup$ @player3236 Maybe, maybe not. In my time group actions were (frequently) used to obtain the class equation. But, after normal subgroups were treated. $\endgroup$ Sep 8 '20 at 15:01
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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, $H⊂Z(G)$

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  • $\begingroup$ Thanks! This answer makes a lot of sense. I realize now that the order of each class must divide the order of the group. I take it this is a given but a conjugacy class can contain p elements if the conjugacy class contains the identity right? Because then ghg^-1 = e means h is e, which is obviously in Z(G). If we take h not equal to e, then the class can contain at most (p-1) elements. $\endgroup$
    – John Wick
    Sep 8 '20 at 16:41
  • $\begingroup$ No, class of identity element contains only identity, see $geg^{-1}=(ge)g^{-1}=gg^{-1}=e $ , so, that's why I say each class can contain at most $(p-1)$ elements as $|H|=p$. $\endgroup$
    – A learner
    Sep 8 '20 at 16:50
  • $\begingroup$ And, also, class of non identity element can't contains identity element, clearly, $ghg^{-1}=e \implies (g^{-1}g)h(g^{-1}g)=g^{-1}eg \implies h=g^{-1}g=e $ $\endgroup$
    – A learner
    Sep 8 '20 at 16:54
  • $\begingroup$ Ah ok. I see. Thanks! $\endgroup$
    – John Wick
    Sep 9 '20 at 15:22
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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 $H\cap Z(G)=H$

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    $\begingroup$ I have not yet learned about nilpotency in terms of groups, so I lack the knowledge needed to understand this solution. Thank you for the solution all the same. $\endgroup$
    – John Wick
    Sep 8 '20 at 16:42
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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 $H\le Z(G)$.

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    $\begingroup$ I really appreciate the answer. Although, I think it will be a while before I can understand it. I have not learned most of this stuff yet (or at least the book has not introduced it yet). Thank you all the same. $\endgroup$
    – John Wick
    Sep 8 '20 at 16:37

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