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perfect group

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Definition

A group GG is perfect if it equals 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}

Examples

The trivial group is perfect, trivially.

Proposition

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

Proposition

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:

Proposition

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(/(2))SL_2(\mathbb{Z}/(2)) and SL 2(/(3))SL_2(\mathbb{Z}/(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.

Proposition

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.

Remark

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

Proposition

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

Proof

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).

References

Last revised on October 28, 2018 at 09:39:50. See the history of this page for a list of all contributions to it.