# 5.3: Classifying Finite Groups

- Page ID
- 699

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We've seen that group theory can't distinguish between groups that are isomorphic. So a natural question is whether we can make a list of all of the groups!

We can make new groups from old groups using the direct product. So it would be nice to focus on groups that are **not** direct products. In the commutative case, this turns out to be pretty straightforward: a (finite) commutative group is a direct product of subgroups if and only if it has a proper subgroup.

The non-commutative case is much more difficult, though. There are actually a few other ways to build new groups from old groups; the most important of these other ways is the *semi-direct product*; we won't describe how to build semi-direct products here, but you can read about them elsewhere. Importantly, one can 'undo' a semi-direct product using a quotient, the same way one can undo a direct product. To get a sense of how useful the construction is, the symmetric group \(S_n\) is the semi-direct product of \(A_n\) and \(\mathbb{Z}_2\). Also, the dihedral group \(D_n\) is a semi-direct product of \(\mathbb{Z}_n\) and \(\mathbb{Z}_2\).

An interesting question, then, is 'Which groups have no quotients?' We've seen that we can form a quotient group whenever there is a normal subgroup.

Definition 5.3.0: Simple Groups

A group is *simple* if it has no proper normal subgroups. (A proper subgroup is any subgroup of \(G\) that is not equal to \(G\) or \(\{1\}\), which are always normal subgroups.)

We'll now actually classify all of the finite simple groups, and discuss some of the history of the non-commutative case.

### The Commutative Case

We can actually classify all of the finite commutative groups pretty easily. First, recall that *every* subgroup of a commutative group is normal.

Proposition 5.3.1 |
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A finite commutative group is simple if and only if it has prime order \(p\). In this case, it is isomorphic to the cyclic group, \(\mathbb{Z}_p\). |

Proof 5.3.2 |
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If a finite commutative group has prime order then it has no proper subgroups, by Lagrange's theorem. Then it must be simple. For the other direction, we assume \(G\) is a finite commutative simple group. \(G\) must be cyclic, or else we could form a proper subgroup by taking powers of a generator. So \(G\sim \mathbb{Z}_n\) for some \(n\). But if \(n\) is not prime we can find a subgroup using a proper divisor of \(n\). Then \(G\sim \mathbb{Z}_p\) for some prime \(p\). |

Theorem 5.3.3 |
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Every finite commutative group is a direct product of cyclic groups of prime order. |

Proof 5.3.4 |
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Let \(A\) be a commutative group with \(n\) elements. Take any element \(x\) not equal to the identity in \(A\); we know that there is some minimal integer \(m\) for which \(x^m=1\). Then \(A\) has a subgroup of order \(m\) generated by \(x\), isomorphic to \(\mathbb{Z}_m\). As a result, we have \(A\sim A_1 \otimes \mathbb{Z}_m\), where \(A_1\) is the quotient \(A/ \mathord \mathbb{Z}_m\). We can repeat that procedure indefinitely (taking an \(x\) in \(A_1\) and writing \(A_1\) as a product, and so on), until we obtain a decomposition \(A=\mathbb{Z}_{m_1}\otimes \mathbb{Z}_{m_k}\), a product of cyclic groups. We can then use the same trick to decompose each \(\mathbb{Z}_m\) into a direct product of cyclic groups of prime order, completing the proof. |

One can extend this trick to some infinite groups: those which have a finite number of generators. (Such groups, unsurprisingly, are called *finitely-generated*.) This gives rise to the *Fundamental theorem of finitely-generated commutative groups*.

Exercise 5.3.5

Suppose \(A\) is a finitely generated commutative group with infinite cardinality. Show that \(A\sim \mathbb{Z} \otimes A'\), where \(A'\) is a finitely-generated commutative group.

### The Non-Commutative Case

One of the major projects of 20th century mathematics research was to classify all of the finite simple groups; the project took fifty years, and the proof of the classification is estimated to span 10,000 pages written by over 100 authors. There's currently an effort underway to simplify the proof, however.

The classification shows that all finite simple groups are of one of four types:

- Commutative groups of prime order,
- Alternating groups \(A_n\) with \(n\geq 5\),
- Groups of Lie type,
- The 26 sporadic groups.

We've already seen the first two types of simple group. It turns out that 'most' finite simple groups are in the third class, groups of Lie type, which are well beyond the scope of these notes to construct. Basically, though, groups of Lie type are certain groups of matrices with entries from a *finite field*, which are we'll see in the next chapter. The 'sporadic' groups are just those groups that don't fit into any of the other three classes!

### Contributors

- Tom Denton (Fields Institute/York University in Toronto)