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

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

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

The trivial group is perfect, trivially.

The alternating group $A_5$ is the smallest nontrivial perfect group.

The binary icosahedral group $2I$ 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 $2I \simeq SL_2(\mathbb{F}_5)$ (this prop.), this is a special case of the following class of examples:

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

See for example here, or Lang 02, theorems XIII 8.3 and 9.2. Notice that the smallest of this class of examples is $SL_2(\mathbb{F}_4)$, of order $60$. In a moment we give a simple argument that this example is isomorphic to $A_5$.

A quotient group of a perfect group is again perfect.

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

Given that there are no nontrivial perfect groups of order less than $60$, this proposition shows that $SL_2(\mathbb{F}_4)$ cannot have a proper nontrivial quotient, i.e., it is a simple group. Since $A_5$ is up to isomorphism the only simple group of order $60$, there must be a (non-canonical) isomorphism $A_5 \cong SL_2(\mathbb{F}_4)$.

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 $colim_i G_i$, taking it to $colim_i (G_i)^{ab} \cong \colim_i 1 \cong 1$ (since an arbitrary colimit of initial objects is again initial).

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

- Serge Lang,
*Algebra*, $3^{rd}$ edition, Springer 2002 (pdf)

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