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If $G$ is a finite $p$-group with a nontrivial normal subgroup $H$, then the intersection of $H$ and the center of $G$ is not trivial.

Shaun
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    H is a p-group, so it has a nontrivial center. H is normal, so...? – Qiaochu Yuan Oct 30 '10 at 19:44
  • @Qiaochu: I suspect "its" refers to $G$; that is, $H\cap Z(G)$ nontrivial. – Arturo Magidin Oct 30 '10 at 21:00
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    @Arturo: ah, sorry. The proof I was thinking of actually doesn't work. – Qiaochu Yuan Oct 30 '10 at 21:06
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    There is a nice generalisation of this result. If $G$ is a nilpotent group and $1\neq H\unlhd G$, then $H\cap Z(G)\neq 1$. Since all $p$-groups are nilpotent, your result could be seen as a corollary of this (if you want a different way of looking at things that is). – David Ward May 17 '13 at 08:13

4 Answers4

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Perhaps a slightly less computationally-intensive argument is to simply note that since $H$ is normal, it must be a union of conjugacy classes. Each conjugacy class of $G$ has $p^i$ elements for some $i$; since $H$ contains at least one conjugacy class with $p^0 = 1$ elements (the class of the identity), and $|H|\equiv 0 \pmod{p}$, it must contain other classes with just one element, which must be classes of central elements of $G$.

Arturo Magidin
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  • This is good. :) – BBischof Oct 30 '10 at 22:13
  • This is the same proof as Chandru1's with less notation, isn't it? – HJRW Oct 31 '10 at 06:07
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    @Henry: It's the same idea, without bothering to actually carry out any computations. That's why I said it was "less computationally-intensive". – Arturo Magidin Oct 31 '10 at 15:19
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    I'd rephrase your proof as "As $H$ is normal is $G$, $G$ acts by conjugation on $H$. As $G$ is a $p$-group, the number of fixed points equals the order $|H| \bmod p$ (the length of an orbit equals the index of a point stabilizer, hence all non-trivial orbits have length multiple of $p$) and is therefore divisible by $p$. As $1$ is a fixed points, there has to be another fixed point $h \in H \setminus {1}$ of $G$, and as $h^g = h$ for all $G$, $h \in Z(G)$." – j.p. Oct 31 '10 at 18:03
  • @Jug: that's not a "rephrase", that's a different argument, I would say. – Arturo Magidin Oct 31 '10 at 22:22
  • @Arturo: That's what I thought first, too. But just when I posted my answer, I realized that my orbits are your conjugacy classes, and that my proof is step for step your proof rephrased in a slightly different language. – j.p. Nov 02 '10 at 08:25
  • Dear Arturo! Very nice solution +1. Let me ask you one question regarding your answer: why each conjugacy class of $G$ has $p^i$ elements for some $i$? Cannot get this :( – RFZ Feb 14 '18 at 21:10
  • @RFZ: Because the size of the conjugacy class of $x$ is the index of the centralizer of $x$; the centralizer is a subgroup of $G$, and hence has index $p^i$ for some $i$. – Arturo Magidin Feb 14 '18 at 21:41
  • @ArturoMagidin I'm really sorry for the late question but I'm preparing for an exam. It seems to me that considering the action of $H$ on $H$ by conjugation yields the same proof as all it matters is that the center of $Z(H$) under the action which is $H \cap Z(G)$ is non trivial. But all the proofs I've seen consider the action of $G$ on $H$. Am I blindly wrong somewhere? Thank you! – sthdakot May 04 '18 at 22:57
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    @sthdakot: It is false in general that $Z(H)$ equals $H\cap Z(G)$. The latter is contained in the former, but they need not be equal. For example, the subgroup generated by $i$ i the quaternion group is abelian, so $Z(H)=H$, but $Z(G)\cap H = {1,-1}\neq H$. If you only look at the action of $H$ on itself, you cannot tell what elements lie in the center of $G$. – Arturo Magidin May 05 '18 at 03:23
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    @ArturoMagidin I think my confusion came from the fact that for an element $h \in H$, the centraliser $C_H(h)= H \cap C_N(h)$. This is true, right? – sthdakot May 05 '18 at 17:14
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    @sthdakot: yes, that is true, but if you write it out you will see why this is true but the center assertion is not. $C_H(h) = {x\in H\mid xh=hx}$, and $C_G(h) = {x\in G\mid xh=hx}$. So $x\in C_H(h)\iff x\in C_G(h)\text{ and }x\in H$. However, $Z(H) = {x\in H\mid \forall h\in H(xh=hx)}$, $Z(G) = {x\in G\mid \forall g\in G\mid xg=gx}$; so $x\in Z(H)$ is not equivalent to $x\in Z(G)\text{ and }x\in H$; the latter is a stronger condition ($x$ is required to commute with everything in $G$), and thus satisfied by fewer elements. – Arturo Magidin May 05 '18 at 18:40
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$H$ is normal, consider $G$ acting on $H$ by conjugation.

The class equation yields $$ \left| H \right| = \left| H^G \right| + \sum_i [G:Stab_{h_i}] \, , $$ where $ H^G = \{ h \in H \mid \ ghg^{-1} = h , \ \forall g \in G \} $ are the fixed points and $ Stab_{h_i} = \{ g \in G \ \mid \ gh_ig^{-1} = h_{i} \}\leqslant G $ is the stabilizer of a $h_i \in H$. Observe that in this case $ H \cap Z(G) = H^G $.

$p$ divides $\left| H \right|$ and $[G:Stab_{h_i}]$ for every non trivial orbit, so it divides $\left| H^G \right|$. In particular, $H^G$ is not empty, so there is an element of $H$ that is also in the center of $G$.

Bach
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desijuan
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Let $a_{1}, . . . , a_{k}$ be representatives of the conjugacy classes of $G$, ordered such that $a_{m} \in H$ and $a_{m+1}, \cdots , a_{k} \notin H$. The conjugacy class $C(a_{i})$ have either $C(a_{i}) \subset H$ or $C(a_{i}) \cap H = \{e\}$. First arrange the $\{a_{1}, . . . , a_{m}\}$ so that the first $r$ represent conjugacy classes of size 1, (i.e. elements in $H \cap Z$) and the latter $m − r$ represent classes of size larger than 1. Then we can write the class equation for $H \cap Z = H$ as: $$|H| = \sum\limits_{i=1}^{m} |C(a_{i}) \cap H| = |H \cap Z| + \sum\limits_{i=r}^{m} |C(a_{i})| = |H \cap Z| + \sum\limits_{i=r}^{m}\frac{|G|}{|N(a_{i})|}$$

As $|H| < p^{n}$ every term in the sum is divisible by $p$ so $|H \cap Z|$ from above is divisible by $p$.

Frenzy Li
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Can we try induction on $n$ by taking quotients like $G/Z$ WHERE $Z$ IS THE CENTER. More technically it goes like this. We take the quotient group $G/Z$ where $Z$ is the center and as $G$ is a $p-group$,$Z$ is non trivial.Hence $o(G/Z)=o(G)/O(Z)$ is less than $o(G)$ to be ideal for application of induction.Now look at the set $S=(Zx|x\in N)$.My claim is that this set is a subgroup and as a matter of fact a normal subgroup.Because $(Zx_1)(Zx_2)=Zx_1x_2$ and hence the closure and the inverse property is trivial.Again for any $x \in G$ and $n \in N$ $(Zx)(Zn)(Zx)^{-1}=Zxnx^{-1}$ and due to $N$ being normal,that belongs to $S$.Hence $S$ is a normal subgroup in $G/Z$.Now due to non-triviality of $Z$ we can apply induction to claim that the intersection of $S$ with the center of $G/Z$ is non-trivial.That is there is a $a$ not in $Z$ but in $N$ such that $Zax=Zxa$ for all $x \in G$ Hence there is an $a \in N$ such that $axa^{-1}x^{-1}$ belongs to $Z$ for all $x \in G$.Now as $a^{-1}$ belongs to $N$ clearly $xa^{-1}x^{-1}$ be in $N$ and thus $axa^{-1}x^{-1}$ belongs to $N$.But if $N$ and $Z$ has trivial intersection then $ax=xa$ for all $x \in G$ which makes $a \in Z$,a contradiction.

Debam
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