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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.

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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
    
@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 8:13

3 Answers 3

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 $p$.

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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 $G$.

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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 6:07
    
@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
    
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

$H$ is normal, consider $G$ acting on $H$ by conjugation.

The class ecuation yields: $$ |H| = \#H^G + \sum\limits_i [H:Stab_{h_i}] $$

Where $ H^G = \{ h \in H \; / \; ghg^{-1} = h \; \; \; \forall g \in G \} $ are the fixed points, and $ Stab_x = \{ g \in G \; / \; gxg^{-1} = x \} $ is the stabilizer of $x$. Observe that in this case $ H \cap Z(G) = H^G$.

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

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