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I have a finite $p$-group $G$ and a normal subgroup $N$ which is not the trivial subgroup. I am asked to show that $|N \cap Z(G)| > 1$.

There has been a similar question on MSE here: How to show that $H \cap Z(G) \neq \{e\}$ when $H$ is a normal subgroup of $G$ with $\lvert H\rvert>1$ Unfortunately, I cannot comment on other user's questions yet, so I made my own question.

I know that the trick is to use conjugation but I can't follow the logic. Can someone elaborate? Especially on the last part of the top answer.

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First, $G$ acts on itself by conjugation $g\cdot x=gxg^{-1}$, and under this action, the orbit $G\cdot x$ is precisely the conjugacy class of $x$ in $G$, by definition essentially. The orbit-stabilizer theorem says $$ |G\cdot x|=\frac{|G|}{|\operatorname{Stab}(x)|} $$ This implies $|\operatorname{Stab}(x)|=\frac{|G|}{|G\cdot x|}$, which means $|G\cdot x|$ divides $|G|$, hence $|G\cdot x|$ is a power of $p$, or possibly $1$.

Since $N$ is normal, it is a union of the conjugacy classes of the elements it contains, and these classes partition $N$. Since $e\in N$, the conjugacy class $C_G(e)=\{e\}\subseteq N$. Since $N$ is nontrivial, it has to contain other conjugacy classes. If all these other classes are not singletons, and since these classes partition $N$, then $|N|=1+\sum\text{powers of }p$, so that $$ 1=|N|-\sum\text{ powers of }p $$ which implies $p\mid 1$, since $p\mid |N|$ and each term which is a power of $p$, a contradiction. (When I say powers of $p$ here, I'm not counting $p^0=1$, since I assumed all the other conjugacy classes are not singletons.) So $N$ must contain some other conjugacy class which is a singleton, say some nonidentity $x\in N$ such that $C_G(x)=\{x\}$. This means $gxg^{-1}=x$ for all $g\in G$, or $gx=xg$ for all $g\in G$, or $x\in Z(G)$, and you conclude.

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    $\begingroup$ Thanks, I think this clears it up. I try to write it down myself and come back in case I encounter any difficulties. $\endgroup$ – Marc Nov 24 '14 at 6:13
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  1. Each $a \in G$ belongs to a conjugacy class, and in a given conjugacy class each element is a conjugate of every other element. G is partitioned into separate conjugacy classes (this is because "$a$ is conjugate to $b$" is an equivalence relation on G).

  2. The orbit-stabilizer theorem shows that the size of each conjugacy class is a power of $p$. Why? Because given $g \in X$ in a conjugacy class $X$, let $Z_g$ be the subgroup of elements fixing $g$ by conjugation (i.e. they commute with $g$). There's a natural bijection between cosets of $Z_g$ and members of $X$. Since $G$ is a $p$-group, the order of $Z_g$ and therefore the index of $Z_g$, which is $|X|$, must be powers of $p$.

  3. If $a \in Z(G)$, the conjugacy class of $a$ is just $\{a\}$, of order $p^0 = 1$. Otherwise the size of the conjugacy class of $a$ is a multiple of $p$.

  4. $N$ is normal, so if $a \in N$, $a$ cannot get out of $N$ by conjugation. It follows that if $N$ includes some element of some conjugacy class, it includes the entire class. In other words, $N$ is a union of some conjugacy classes.

  5. $N$ already includes one conjugacy class of size $1$: that of $e$. If it didn't have any other elements of $Z(G)$ in it, all other conjugacy classes in $N$ would be of size that divides $p$. But then the total sum of all these orders of conjugacy classes in $N$ would be $1$ modulo $p$. This cannot be, as the order of $N$ must divide $|G|$.

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  • $\begingroup$ Thanks, Avva! I am going to write it down for myself now and see if I run into trouble. $\endgroup$ – Marc Nov 24 '14 at 6:16
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The same question was asked and closed here, where the OP asks for a proof using the upper central series of $G$.


Define $i$ as the smallest integer such that $Z_i\cap N\neq\lbrace e\rbrace$, and consider $g\in (Z_i\cap N)\setminus\lbrace e\rbrace$. Since $Z_0=\lbrace e\rbrace$ and for $n$ large enough $Z_n=G$, $i$ is well defined and $\geq 1$. Then for any $h\in G$, $$g\underbrace{hg^{-1}h^{-1}}_{\in N}\in Z_{i-1}\cap N$$ i.e. $\forall h\in G, ghg^{-1}h^{-1}=e$ i.e. $g\in Z(G)$. Hence $N\cap Z(G)\neq\lbrace e\rbrace$

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  • $\begingroup$ How did you infer that last line? $\endgroup$ – Meitar Jan 17 '15 at 17:02
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    $\begingroup$ This is because, by definition of $i$, $Z_{i-1}\cap N=\lbrace e\rbrace$. $\endgroup$ – Olivier Bégassat Jan 17 '15 at 17:21

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