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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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    $\begingroup$ H is a p-group, so it has a nontrivial center. H is normal, so...? $\endgroup$ – Qiaochu Yuan Oct 30 '10 at 19:44
  • $\begingroup$ @Qiaochu: I suspect "its" refers to $G$; that is, $H\cap Z(G)$ nontrivial. $\endgroup$ – Arturo Magidin Oct 30 '10 at 21:00
  • $\begingroup$ @Arturo: ah, sorry. The proof I was thinking of actually doesn't work. $\endgroup$ – Qiaochu Yuan Oct 30 '10 at 21:06
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    $\begingroup$ 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). $\endgroup$ – David Ward May 17 '13 at 8:13
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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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  • $\begingroup$ This is good. :) $\endgroup$ – BBischof Oct 30 '10 at 22:13
  • $\begingroup$ This is the same proof as Chandru1's with less notation, isn't it? $\endgroup$ – HJRW Oct 31 '10 at 6:07
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    $\begingroup$ @Henry: It's the same idea, without bothering to actually carry out any computations. That's why I said it was "less computationally-intensive". $\endgroup$ – Arturo Magidin Oct 31 '10 at 15:19
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    $\begingroup$ @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 $G$. $\endgroup$ – Arturo Magidin May 5 '18 at 3:23
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    $\begingroup$ @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. $\endgroup$ – Arturo Magidin May 5 '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 $G$.

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  • $\begingroup$ Can someone explain to me this symbol? I know what stabilizer is, I just don't get what $[H:Stab_{h_i}]$ is. $\endgroup$ – I_Really_Want_To_Heal_Myself Jan 17 '17 at 16:30
  • $\begingroup$ @I_Really_Want_To_Heal_Myself: The index of the stabilizer in $H$, which is defined to be the number of cosets produced when partitioning $H$ into cosets of the stabilizer. $\endgroup$ – Jacob Manaker Jun 17 '18 at 4:44
  • $\begingroup$ @I_Really_Want_To_Heal_Myself Actually, it is $ [G:Stab_{h_i}] $. $\endgroup$ – Bach Mar 26 at 13:41
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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 $p$.

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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.

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