prove: for any group , there exist a unique maximal perfect subgroup of $G$ and that this subgroup is normal

this is an exercise in Dummit and Foote , 3rd ed

but i really think that this statement is not accurate !

for example , suppose that $G=Z_p$ for some prime $p$ , then $G$ has no proper subgroups so it doesn't have any maximal subgroup ! so the statement is false

also suppose that $G$ is simple group and suppose that we were able to construct such maximal perfect subgroup but we are sure that this subgroup is not normal since $G$ is simple so the statement is again false !

so , am i misunderstanding the exercise or is it really not accurate ?


Clear definition

The statement is accurate, but must be understood correctly: in every group $G$ there is a subgroup $D \leq G$ with the following properties: (1) $D$ is perfect, (2) if $H \leq G$ and $H$ is perfect, then $H \leq D$, and (3) $D$ is normal.

By (1) and (2) there is only one such subgroup, so (3) is not surprising.


If $G$ is cyclic of prime order (or abelian, or solvable), then $D=1$ is the trivial subgroup. This is the only perfect subgroup of $G$, so it satisfies (2) in an unimpressive way.

If $G$ is non-abelian simple, then $G$ itself is perfect, so $G \leq D$ by (2), but since $D \leq G$, we get $D=G$.

Finding $D$

If $G$ is finite, then one can find $D$ using Jared's Hint #1 and Maths Lover's observation in the comments: If $H \leq G$ then $[H,H] \leq [G,G]$. Notice that $D=[D,D] \leq [G,G]$ and when we try to find the $D$ for $G$, we would get the same as if we tried to find the $D$ for $[G,G]$. Hence we just keep replacing $G$ with $[G,G]$, until finally $G$ and $[G,G]$ are equal. At that point they are equal to $D$. This stops after a finite number of steps in a finite group. [ For an infinite group the same idea works, but at limit ordinals you take intersections, and the overall process is transfinite induction. ] This is called the “residual” method, and writes $D$ as the solvable [ or hypoabelian ] residual.

However, Dummit–Foote outline a different method, the radical method. If $A,B$ are perfect subgroups then set $C=\langle A,B \rangle$. Then $C = \langle A , B \rangle = \langle [A,A], [B,B] \rangle \leq [C,C] \leq C$, so $C=[C,C]$ is also perfect. $D = \langle A \leq G : A = [A,A] \rangle$ is thus a perfect subgroup of $G$.

  • $\begingroup$ how did you know that , $ \langle [A,A], [B,B] \rangle \leq [C,C]$, ? $\endgroup$ – Fawzy Hegab Jun 13 '13 at 14:51
  • $\begingroup$ $A\leq C$ so $[A,A] \leq [C,C]$ $\endgroup$ – Jack Schmidt Jun 13 '13 at 14:59
  • $\begingroup$ this is so nice way of proving ! i used the definitions and worked through them to prove that $C$ is also perfect , but using the fact that $X^*=[X,X]$ where $X^{*}$ is the commutator subgroup of $X$ as you have done is so wonderful ! thanx so much :) $\endgroup$ – Fawzy Hegab Jun 13 '13 at 15:04
  • $\begingroup$ Resemble to part of a concrete paper (+1) $\endgroup$ – mrs Jun 13 '13 at 16:36

Hint 1: For finite groups, consider the last term of the derived series of the group, which is allowed to be trivial!

Hint 2: For arbitrary groups, can you show that a subgroup generated by perfect subgroups is itself perfect?

  • $\begingroup$ the derived series is provided in the next chapter , so i should solve it without using it . also my question doesn't ask for a hint but about the understanding of the statement itself as i have showed in the question $\endgroup$ – Fawzy Hegab Jun 13 '13 at 7:03
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    $\begingroup$ @MathsLover: I did address your understanding, although not explicitly I guess. When I say the subgroup is allowed to be trivial, what I mean is that this subgroup you are looking for may consist solely of the identity element. In the case of $\mathbb{Z}/p\mathbb{Z}$, the trivial subgroup is a maximal subgroup, and in the case of a simple group the trivial subgroup is normal. $\endgroup$ – Jared Jun 13 '13 at 7:06
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    $\begingroup$ @MathsLover: I think your intuition is leading to the following: in the case of simple groups, we should expect this unique maximal perfect subgroup to be the trivial subgroup. $\endgroup$ – Jared Jun 13 '13 at 7:14
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    $\begingroup$ @MathsLover: Ah yes, you're right. A maximal subgroup is a proper subgroup that is strictly contained in no other proper subgroup. In $\mathbb{Z}/p\mathbb{Z}$, strangely enough, the trivial subgroup satisfies this definition. $\endgroup$ – Jared Jun 13 '13 at 7:27
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    $\begingroup$ "Maximal perfect subgroup" means "a perfect subgroup that contains all other perfect subgroups". It is very rarely a maximal subgroup of the group. If the group itself is non-abelian simple, then the group itself is the maximal perfect subgroup. $\endgroup$ – Jack Schmidt Jun 13 '13 at 11:26

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