# 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.

• I believe there's a typo here: "Prove that if H is a subgroup ... then H is a subgroup of G." Sep 8 '20 at 14:46
• @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.
– ΑΘΩ
Sep 8 '20 at 14:53
• 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 $H$?) Sep 8 '20 at 14:58
• @DanielFischer If OP hasn't learnt about normal subgroups, they probably haven't learnt group actions. Sep 8 '20 at 14:59
• @player3236 Maybe, maybe not. In my time group actions were (frequently) used to obtain the class equation. But, after normal subgroups were treated. Sep 8 '20 at 15:01

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, $$H⊂Z(G)$$

• 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. Sep 8 '20 at 16:41
• 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 $|H|=p$. Sep 8 '20 at 16:50
• And, also, class of non identity element can't contains identity element, clearly, $ghg^{-1}=e \implies (g^{-1}g)h(g^{-1}g)=g^{-1}eg \implies h=g^{-1}g=e$ Sep 8 '20 at 16:54
• Ah ok. I see. Thanks! Sep 9 '20 at 15:22

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 $$H\cap Z(G)=H$$

• 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. Sep 8 '20 at 16:42

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 $$H\le Z(G)$$.

• 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. Sep 8 '20 at 16:37