Simple Group

A simple group is a group with no nontrivial proper normal subgroups. The Jordan–Hölder theorem gives a recipe for breaking a finite group down as a certain combination of simple groups. So in a sense, finite simple groups are the building blocks of finite group theory.

One of the largest and most ambitious mathematical research projects of the late $20^\text{th}$ century was the classification of all finite simple groups, which was completed in 2008 (though most of the work was finished by the mid-1980s). The theory of finite simple groups includes several quite beautiful examples coming from fields such as combinatorics and geometry, including the famous monster group, which has more than $10^{53}$ elements, and has deep connections to modern number theory and even string theory, via the theory of modular forms.

A simple group is a nontrivial group $G$ such that if $N \trianglelefteq G$ is normal, then $N = \{1\}$ or $N=G.$

The group ${\mathbb Z}_p$ of integers mod $p$ is simple, for $p$ a positive prime number. This is clear by Lagrange's theorem, since the order of a subgroup $N$ divides $p,$ so it is either $1$ or $p.$ So ${\mathbb Z}_p$ has no nontrivial proper subgroups (since it is abelian, all subgroups are automatically normal).

Any finite abelian simple group is isomorphic to ${\mathbb Z}_p$ for some prime $p.$

Let $x$ be a non-identity element of a finite abelian simple group $G.$ Then the subgroup generated by $x$ is normal (all subgroups of an abelian group are automatically normal), so it must be all of $G.$ So $G$ is a cyclic group generated by $x$ (that is, $G$ consists of the powers of $x$).

If $|G|=n,$ then the homomorphism $f \colon {\mathbb Z}_n \to G$ given by $f(k) = x^k$ is clearly an isomorphism, so $G$ is isomorphic to ${\mathbb Z}_n.$

Since $G$ is generated by any non-identity element, so is ${\mathbb Z}_n.$ This can only be true if $n$ is prime--if $n$ were composite, suppose $d|n$ with $d$ and $n/d$ not equal to $1.$ Then the subgroup $d{\mathbb Z}_n$ is a nontrivial proper normal subgroup, since $d{\mathbb Z}_n$ consists of the $n/d$ elements mod $n$ which are multiples of $d.$ $($Equivalently, in $G,$ $x^d$ does not generate $G,$ because the subgroup generated by $x^d$ consists of the $n/d$ elements $x^a,$ where $d|a.)$

The most accessible non-abelian simple group is the alternating group $A_5.$ Its simplicity was discovered by the great (and tragically short-lived) French mathematician Galois, and is related to the fact (due to Abel and Ruffini) that there is no way to solve a quintic polynomial equation "by radicals," i.e. there is no expression using only algebraic operations $($including $n^\text{th}$ roots$)$ on the coefficients of the polynomial that produces a general solution to a quintic. (Put more succinctly, there is a quadratic formula, and much more difficult cubic and quartic formulas; but there is no quintic formula.)

Facts about $A_n$:

• $A_n$ is a subgroup of the symmetric group $S_n$ of permutations on $n$ symbols. It consists of the even permutations, which are products of an even number of transpositions.
• $A_n$ is the kernel of the sign homomorphism $\epsilon \colon S_n \to \{\pm 1\}.$ It is a normal subgroup of $S_n$ of index $2,$ so its order is $\frac{n!}{2}.$
• $A_n$ is simple for $n \ge 5.$ $($This implies that there are no sextic, septic, ... formulas for the roots of a polynomial of degree $\ge 5$; see the below section on solvability.$)$

Here is a collection of classical theorems about finite simple groups. These served as motivation for the general classification of all finite simple groups (see below).

Let $G$ be a finite nonabelian simple group. Then

• $|G|$ must be divisible by at least three distinct primes. (Burnside, 1904)

• $|G|$ must be even. (Feit-Thompson, 1963)

• Let $p$ be a prime dividing $|G|.$ Let $D$ be the set of divisors of $|G|$ which are congruent to $1 \pmod p.$ If the only element of $D$ is $1,$ then $G$ has a normal subgroup of order $p^k$ for some $k,$ and is hence not simple. (Sylow, 1872)

There are $52$ positive integers $\le 1000$ which pass the tests in the above theorem as possible orders of nonabelian simple groups. The list begins $30,60,90,120,132,150,168,\ldots$. The smallest two numbers on the list which are actually orders of simple groups are $60 = |A_5|$ and $168,$ which is the order of a group $PSL(2,7)$ called a projective special linear group.

There are other tricks that can restrict the search for finite simple groups of a given order; the following exercise indicates one of them. Recall that the index of a subgroup $H$ of a finite group $G$ is $|G|/|H|.$

Let $G$ be a finite group and let $H$ be a subgroup of index $n.$ Show that $G$ has a normal subgroup of index $\le n!$ contained in $H.$

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Consider the $n$ cosets $C_1 = a_1H, C_2 = a_2H, \ldots, C_n = a_nH,$ where the $a_i$ are representative elements of each coset. Left multiplication by an element $g \in G$ permutes these cosets, so each element of $G$ corresponds to a permutation of the cosets, and hence an element of $S_n.$ This gives a function $\phi \colon G \to S_n,$ and it is straightforward to show that this is actually a homomorphism $($this boils down to the statement that multiplying on the left by $g_1$ and then by $g_2$ is the same as multiplying by $g_2g_1).$

By the first isomorphism theorem, the map $\phi \colon G\to S_n$ gives an isomorphism of $G/\text{ker}(\phi)$ onto its image, which is a subgroup of $S_n.$ Let $N = \text{ker}(\phi)$; then $N$ is normal and its index equals the size of its image, which is $\le n!$. Note also that multiplying by an element of $N$ fixes every coset, so it must fix the trivial coset $H.$ So $nH = H$ for any $n \in N,$ which shows that $n \in H.$ So $N$ is contained in $H.$ $_\square$

This fact, coupled with the Sylow theorems, can sometimes be used to rule out simple groups of certain orders as well.

Every finite group $G$ has a series of subgroups of the form $\{1\} \triangleleft H_1 \triangleleft H_2 \triangleleft \cdots \triangleleft H_{k-1} \triangleleft H_k = G,$ where each $H_i$ is a maximal normal subgroup of $H_{i+1}.$ This condition is equivalent to the statement that the quotient groups $H_{i+1}/H_i$ are simple, by the third isomorphism theorem. This series is called a composition series. The simple quotient groups are called the composition factors.

If $G=S_4,$ one composition series for $G$ is as follows: $\{ 1 \} \triangleleft {\mathbb Z}_2 \triangleleft V_4 \triangleleft A_4 \triangleleft S_4.$ Here $V_4$ is the group $\big\{i,(12)(34),(13)(24),(14)(23)\big\}$ $\big($and ${\mathbb Z}_2$ is the subgroup generated by, say, $(12)(34)\big).$

If $G=S_5,$ there is a simpler composition series: $\{1 \} \triangleleft A_5 \triangleleft S_5.$

A finite group $G$ is called solvable if it has a composition series whose composition factors are all isomorphic to ${\mathbb Z}_p$ for some $p.$

The terminology comes from the solving of polynomial equations: it turns out that there is a formula for the roots of a polynomial of degree $n$ using only $k^\text{th}$ roots and algebraic expressions in the coefficients if and only if $S_n$ is solvable.

(Jordan–Hölder) Every composition series of a finite group $G$ has the same composition factors (up to permutation).

A corollary of the Jordan–Hölder theorem is that $S_n$ is not solvable for $n \ge 5,$ because $A_n$ is simple and the composition series $1 \triangleleft A_n \triangleleft S_n$ has a composition factor that is not a cyclic group of prime order.

The complete list of finite simple groups was completed in 2008, although the vast majority of the work on the classification problem was done in the 1970s and early '80s. The full proof currently spans tens of thousands of pages, and is in the process of being refined into a more streamlined form. Here is a partial description of the classification.

A finite simple group $G$ is isomorphic to one of the following groups:

• ${\mathbb Z}_p,$ for some prime $p$
• $A_n,$ for some $n \ge 5$
• a simple group "of Lie type," which is a member of one of several infinite families of groups related to groups of linear transformations over a finite field
• one of 26 "sporadic" groups that do not fall into any of the above families.

The monster group mentioned in the introduction is the largest of the sporadic groups. It can be written down as a certain subgroup of the group of invertible $196883 \times 196883$ matrices with complex entries, and has order $\approx 8 \times 10^{53}.$