# A question on $p$-groups, and order of its commutator subgroup.

$\textbf{QUESTION-}$ Let $P$ be a p-group with $|P:Z(P)|\leq p^n$. Show that $|P'| \leq p^{n(n-1)/2}$.

If $P=Z(P)$ it is true. Now let $n > 1$, then

If I see $P$ as a nilpotent group and construct its upper central series, it will end , so let it be,

$e=Z_0<Z_1<Z_2<......<Z_r=P$

Now as $Z_{i+1}/Z_i=Z(P/Z_i)$, so if if I take some $x\in Z_2$\ $Z_1$ then $N$={$[x,y]|y\in P$} $\leq Z_1(P)$ and $N \triangleleft P$, so $P/N$ is a group with order $\leq p^{n-1}$.

Now if I let $H=P/N$ then obviously |$H/Z(H)$|$\leq p^{n-1}$.

Now $H'\cong P'N/N \cong P'/(P' \cap N)$ so from here I could finally bring $P'$ atleast into the picture, now |$P'$|=$|H'|.|P'\cap N|$ so $|P'|\leq |H'||N|$.

This is where I am $\textbf{STUCK}$

Now , from here how can I calculate or find some power $p$ bounds on $|H'|$ and $|N|$ so i could get my result.

• Where does this come from? – Igor Rivin Oct 7 '14 at 19:59
• It is a question in Isaacs' finite group theory – Bhaskar Vashishth Oct 7 '14 at 20:00
• This is not research level... – Igor Rivin Oct 7 '14 at 20:01
• yeah maybe, but nobody on mathstack answered it even after bounty, I though may be somebody here could help me. – Bhaskar Vashishth Oct 7 '14 at 20:04
• OK, here is a hint, which I think works. Let $g \in Z_2(P) \setminus Z(P)$, and let $N=[P,g]$. Then $|N| \le p^{n-1}$ and by induction $|(G/N)'| \le p^{(n-1)(n-2)/2}$. – Derek Holt Oct 7 '14 at 20:33

For $n=0,1$, as you observe, there is nothing to prove.

Suppose now that the proposition is true for $n-1$, $n \geq 2$

The first observation is that if $n \geq 2$, $Z < Z_{2}$, so we can choose $x \in Z_{2} \setminus Z$.

Consider now the function $\alpha: P \rightarrow P$, $y \mapsto \left[ x,y \right]$. Since $x \in Z_{2}$ we have that $\alpha\left( y \right)= \left[ x,y \right] \in Z$ which is abelian so that $\alpha$ is a group morphism. Let $N := \operatorname{Im}\left( \alpha \right)$. We have that $\operatorname{Ker}\left( \alpha \right)= C_{G}\left( x \right)$ (the centraliser of $x$) and that $Z < C_{G}\left( x \right)$ since $x \in C_{G}\left( x \right) \setminus Z$. From this it follows

$$\left| N \right| = \left| \dfrac{P}{C_{G}\left( x \right)} \right| < \left| \dfrac{P}{Z} \right| \leq p^n$$

i.e.

$$\tag{1}\label{1} \left| N \right| \leq p^{n-1}$$

$N$ is normal in $P$ because it is central and we can construct $H = P/N$. We have that $\dfrac{Z}{N} < Z\left( H \right)$ because $xN \in Z\left( H \right) \setminus \dfrac{Z}{N}$ so that

$$\left| \dfrac{H}{Z\left( H \right)} \right| < \left| \dfrac{H}{\dfrac{Z}{N}} \right| = \left| \dfrac{P}{Z} \right| \leq p^n$$

we can that deduce that $\left| H / Z\left( H \right) \right| \leq p^{n-1}$ and, by induction hypothesis,

$$\tag{2}\label{2} \left| H^\prime \right| \leq p^{\frac{\left( n-1 \right) \left( n-2 \right)}{2}}$$.

The final observation is that $N \leq P^\prime$ and

$$\tag{3}\label{3} P^\prime / N \leq H^\prime$$ and putting all together we have:

$$\left| P^\prime \right| \stackrel{\eqref{3}}\leq \left| H^\prime \right| \left| N \right| \stackrel{\eqref{1}\eqref{2}}\leq p^{\frac{\left( n-1 \right) \left( n-2 \right)}{2}} p^{n-1} = p^{\frac{n \left( n-1 \right)}{2}}$$

• one doubt.. why $\alpha\left( y \right)= \left[ x,y \right] \in Z$ – Bhaskar Vashishth Nov 27 '14 at 22:16
• @BhaskarVashishth Because $x \in \operatorname{Z}_2$ iff $xZ \in \operatorname{Z}\left(\frac{P}{Z}\right)$ iff $\forall y \in P \; x\operatorname{Z}y\operatorname{Z}=y\operatorname{Z}x\operatorname{Z}$ iff $\forall y \in P \; [x,y] \in \operatorname{Z}$. PS I'm glad you liked my answer because I gave it to your first post (where there were no answers) and it took me a good afternoon of work, and I was very happy.Then I saw the post on MO and that D. Holt, the day before, had already given the same strategy of solution, and I felt horribly.But in the end it seems that my work wasn't worthless – Giuliano Bianco Nov 27 '14 at 23:15
• Does $Z_{2}/Z_1=Z(P/Z_1)$ also says $Z_{2}/Z=Z(P/Z)$. Because $Z_{2}/Z_1=Z(P/Z_1)$ is we get by definition of central series. I guess it is because $Z_{2}/Z \subseteq Z_{2}/Z_1$ and $Z(P/Z) \subseteq Z(P/Z_1)$. Right. Always a basic doubt... And hard work always pays off. Thanks for this comprehensible answer... – Bhaskar Vashishth Nov 27 '14 at 23:25
• @BhaskarVashishth I'm used to the numbering starting with $Z_0=1$, $Z_1=Z=\operatorname{Z}\left(P\right)$ so by $Z_2$ I mean that $\frac{Z_2}{Z}=\operatorname{Z}\left(\frac{P}{Z}\right)$ – Giuliano Bianco Nov 27 '14 at 23:32
• ok ok sorry ... its all clear now. thanks a lot – Bhaskar Vashishth Nov 27 '14 at 23:34

The case when $n=1$ is trivial, since a group $P$ such that $P/Z(P)$ is cyclic is already abelian.

Now consider a maximal subgroup $M$ containing $Z(P)$, where $[P:M]=p$. Then by induction (since $Z(P)\le Z(M)$) $|M'|\le p^{(n-1)(n-2)/2}$.

If we look at $P/M'$, then $M/M'$ is an abelian normal subgroup such that (by the second isomorphism theorem) $(P/M')/(M/M')$ is cyclic (since $P/M$ is).

By Lemma 4.6 in FGT $|M/M'| = |P'/M'|\cdot |M/M'\cap Z(P/M')|$.

This is the same as $|M| =|P'|\cdot |M/M'\cap Z(P/M')|$.

Let $K/M' = Z(P/M')$. Thus we have $$|M| = |P'|\cdot \dfrac{|M\cap K|}{|M'|}.$$

If we can show $Z(P)\le M\cap K$, then we have $$|M| \ge |P'|\cdot\dfrac{|Z(P)|}{|M'|},$$

which is equivalent to $$|P'| \le \dfrac{|M|\cdot |M'|}{|Z(P)|},$$

which implies $$|P'| \le p^{n(n-1)/2}$$

by the bound on $|M'|$, the bound on $|P/Z(P)|$, and the fact $M$ is a maximal subgroup (of index $p$).

So we already know $Z(P)\le M$ (by definition), and since the image of $Z(P)$ in $P/M'$ is central, we have $Z(P)\le K$. We are done.

• @GiulianoBianco: Both $M$ and $K$ contain $M'$, so this is just the correspondence (fourth isomorphism) theorem. – Guest Oct 13 '14 at 17:30