nLab perfect group




A (discrete) group GG is perfect if it is equal to its own commutator subgroup [G,G][G, G], i.e., if every element of GG is a product of commutators (elements of the form [g,h]=ghg 1h 1[g, h] = g h g^{-1} h^{-1}).

Equivalently: let G abG^{ab} denote the abelianization of GG (the target of the homomorphism GG abG \to G^{ab} that is universal among homomorphisms from GG to abelian groups, or the largest abelian quotient of GG). Then GG is perfect precisely when G abG^{ab} is a trivial group, since G abG/[G,G]G^{ab} \cong G/[G, G]:

GperfectG abis trivial. G \;\; \text{perfect} \;\;\;\Leftrightarrow\;\;\; G^{ab} \;\; \text{is trivial} \,.



The trivial group is perfect, trivially.


The alternating group A 5A_5 is the smallest nontrivial perfect group.


The binary icosahedral group 2I2I is a perfect group: its abelianization is the trivial group.

In fact, up to isomorphism, the binary icosahedral group is the unique finite group of order 120 which is a perfect group.

Since 2ISL 2(𝔽 5)2I \simeq SL_2(\mathbb{F}_5) (this prop.), this is a special case of the following class of examples:


The special linear group SL n(𝔽)SL_n(\mathbb{F}) is perfect for any field 𝔽\mathbb{F} and any n1n \geq 1, except for the cases SL 2(𝔽 2)SL_2(\mathbb{F}_2) and SL 2(𝔽 3)SL_2(\mathbb{F}_3).

See for example here, or Lang 02, theorems XIII 8.3 and 9.2. Notice that the smallest of this class of examples is SL 2(𝔽 4)SL_2(\mathbb{F}_4), of order 6060. In a moment we give a simple argument that this example is isomorphic to A 5A_5.


A quotient group of a perfect group is again perfect.

This last assertion is easy to see: GG is perfect if it has no nontrivial abelian quotients. If a quotient HH had a nontrivial abelian quotient, then obviously so would GG.


Given that there are no nontrivial perfect groups of order less than 6060, this proposition shows that SL 2(𝔽 4)SL_2(\mathbb{F}_4) cannot have a proper nontrivial quotient, i.e., it is a simple group. Since A 5A_5 is up to isomorphism the only simple group of order 6060, there must be a (non-canonical) isomorphism A 5SL 2(𝔽 4)A_5 \cong SL_2(\mathbb{F}_4).

Relatedly, we have


An arbitrary colimit of perfect groups (as calculated in Grp, the category of groups) is again perfect.


The abelianization functor, being a left adjoint, preserves a colimit of perfect groups colim iG icolim_i G_i, taking it to colim i(G i) abcolim i11colim_i (G_i)^{ab} \cong \colim_i 1 \cong 1 (since an arbitrary colimit of initial objects is again initial).


If the terminal map BG*\mathbf{B}G \to \ast out of the delooping groupoid BG\mathbf{B}G of a discrete group GG is an epimorphism in the \infty -category Grpd Grpd_\infty of \infty -groupoids, then GG must be perfect (see the discussion there).


Last revised on June 12, 2022 at 20:38:33. See the history of this page for a list of all contributions to it.