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I want to Show that if $G$ is a finite group and if $[G:Z(G)]=n$, then $G' $ has at most $n^2$ elements.

To make it clear, I will define some of the terms and notations used above:

  • $Z(G)$ is the center of the group G.

  • $[G:Z(G)] = \dfrac{\vert G \vert}{\vert Z(G) \vert} $

  • $G'$ is the derived subgroup i.e. $G'=\langle\{[g,h] \mid g,h \in G \}\rangle$

Now I'm having trouble to link those 2 elements ($[G:Z(G)]$ and $G'$ that is).

Could I have a hind or a clarification as to how to relate the order of $G'$ and $[G:Z(G)]$?

Edit Since I am having trouble answering the question, Where could I have a proof of that statement?

Sylvester Stallone
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  • The (very unspecific) relation between the two cardinalities is that if a lot of elements commute (which tends to make $Z(G)$ big), then there are few commutators (i.e. a lot of the $[g,h]$ end up being the identity element, and a lot of the others end up being equal to each other). – Arthur Oct 25 '16 at 20:02
  • should $G$ be finite? – M. Van Oct 25 '16 at 20:10
  • I don't understand why if $Z(G)$ is big then $[g,h]$ is small. Isn't $\vert [g,h] \vert$ only determined by the size of G? – Sylvester Stallone Oct 25 '16 at 20:14
  • @M.Van Yes! I've edited my post – Sylvester Stallone Oct 25 '16 at 20:15
  • A small note - $|G|$ is known as the order of the group, not the cardinality. – Mark Schultz-Wu Oct 25 '16 at 20:17
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    @Mark Both are used, and for groups there is no difference. – flawr Oct 25 '16 at 20:19
  • For instance, if $G=A_5$, the alternating group on five elements, where a lot of elements do not commute, $G'$ is the whole group. In the other end of the scale, if $G$ is abelian, then $G'={e}$. Note that the commutator $[g,h]=ghg^{-1}h^{-1}$, which is $e$ iff $g$ and $h$ commute. So, the more elements in a group commute, the more pairs give a trivial commutator. – Arthur Oct 25 '16 at 20:47
  • The statement is not true so you will not be able to prove it. Where does the question come from? – Derek Holt Oct 25 '16 at 21:04
  • @DerekHolt Could you provide a counter-example? It comes from an exercice sheet from my uni – Sylvester Stallone Oct 25 '16 at 21:06

3 Answers3

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Obviously, the number of commutators, and not the whole commutator subgroup, is bounded by $n^2$, since if $u, v \in Z(G)$ and $x,y \in G$, then $[ux,vy]=[x,y]$, so the number of commutators is only depending on the number of cosets of $Z(G)$.
As for the whole group, it was James Wiegold (Multiplicators and groups with finite central factor groups, Math. Z. 89, 345-347, (1965)) who proved that $|G'| \leq n^{\frac{1}{2}(log_p(n)-1)}$, where $p$ is the smallest prime dividing $n$. This bound is best possible if $n=p^m$ and the exponent becomes equal to $\frac{1}{2}m(m-1)$. See also Derived subgroup where not every element is a commutator

Community
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Nicky Hekster
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If $p$ is a prime and $n>1$ then there are groups of order $p^{n(n+1)/2}$ with $G'=Z(G)$ having order $p^{n(n-1)/2}$. So $|G:Z(G)| = p^n$ but $|Z(G)| >p^{2n}$, and the result is not true.

These groups are nilpotent of class $2$ and have generators $x_1,\ldots,x_n$ all of order $p$, where $G'=Z(G)$ is elementary abelian with generators $[x_i,x_j]$ with $1 \le i < j \le n$.

Derek Holt
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Maybe they should've said the set $S$ of commutators has size $\le n^2$. Note $G'$ may be bigger than $S$.

Hint. Consider the commutator map $[\cdot,\cdot]:G\times G\to G'$.

anon
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