Simple groups

For an arbitrary group, G, we recall that G and e are normal subgroups, if these are the only ones we say the group is simple.

Definition 6.3.1   We say that a group $G \not= \{e\}$ is simple provided that $N \lhd G$ implies $N = G$ or $N = \{e\}$.

One of the most outstanding problems in group theory has been to give a complete classification of all finite simple groups. In other words, this is the program to discover all finite simple groups and to prove that there are no more to be found. This was recently accomplished through the efforts of many mathematicians. The end result of which is called by D. Gorenstein ``The Enormous Theorem,'' as noted in the introduction to this chapter. Certainly one trivial family of finite simple groups would be all groups of prime order $p$, since by Lagrange's Theorem the only subgroups of such groups are of orders $1$ and $p$ (i.e., $\{e\}$ and the whole group). We shall presently determine a less trivial class of simple groups. There are other families of finite simple groups, but their determination is beyond the scope of this book. One of the easiest-to-state results in the proof of the ``Enormous Theorem'' was a tremendous result in itself due to W. Feit and J.G. Thompson (see [FT]). This result took a whole volume of the Pacific Journal of Mathematics (255 pages) to prove; moreover, it is considered to have provided a great deal of impetus to the study of the classification problem. What the Feit-Thompson Theorem (as it is called) basically says (later we shall phrase it in a different form) is that if $G$ is a finite simple group and $\vert G\vert$ is not of prime order then $G$ must be even. This settled an old conjecture of W. Burnside. In the words of D. Gorenstein, ``The complexity of the proof of this easily understood statement (the Feit-Thompson Theorem) foreshadowed the extreme length of the complete classification of the simple groups.''

Let us now turn to the main theorem of this section. First note that $A_3$ is simple, since $\vert A_3\vert = 3!/2 = 3$. As Example 6.1.3 shows, $A_4$ is not simple. It contains a normal subgroup of order 4, viz, the Klein 4-group, $V_4 = \{(1), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}$. However, we have the following result.

Theorem 6.3.2   $A_n$ is simple for $n \geq 5$.

Proof: Let $N \lhd A_n$ where $n \geq 5$. We want to show according to Definition 6.3.1 that if $N\not= \{e\}$, then $N =A_n$. In order to do this, we first show that if $N$ contains a single 3-cycle, i.e., a cycle of length 3, then $N =A_n$. Thus, say $(i,j,k)\in N$. Observe that $(2,i) (1,j)\in A_n$ if $i \not= 2$ and $j \not= 1$, and since $N \lhd A_n$,

$$(2,i) (1,j) (i,j,k) (2,i)^{-1} (1,j)^{-1} \in N,$$

i.e. $(2,1,k)\in N$ (where we have used the fact that disjoint cycles commute). Let $f = (1,2) (k,m)$ where $m \not= 1, 2,k$ but otherwise $m$ is an arbitrary integer less than or equal to $n$. (This can be done since $n \geq 5$.) Then $f \in A_n$ and since $N \lhd A_n$,

$$f(2,1,k)f^{-1} = (1,2,m) \in N$$

also $(1,2,k) = (2,1,k)^2 \in N$. Thus $N$ contains all $(1,2,m)$ for $3 \leq m \leq n$. By exercise 7, Section 3.2, these cycles generate $A_n$, so $N =A_n$, and we are done. If $i = 2$ and $j = 1$, then immediately $(2,1,k)\in N$, and we proceed analogously. If $i = 2$ and $j \not= 1$, then $(2,j,k) \in N$, and so $(1,j) (k,j) (2,j,k) ((1,j) (k,j))^{-1} = (2,k,1) = (1,2,k) \in N$. Then we proceed analogously. Finally if $i \not= 2$ and $j = 1$, a similar argument can be given (see exercise 1 for this section).

Thus to complete the proof, all we must do is show that if $N \lhd A_n$ and $N\not= \{e\}$, then $N$ must contain a 3-cycle. To this end, choose an $f \in N$, $f \not= (1)$, such that $f$ leaves fixed a maximal number of the numbers $1, 2, ..., n$. Suppose $f$ were not a $3$-cycle, then there are just two cases:

Case 1

$f$, in its representation as a product of disjoint cycles, contains a cycle of length $\geq 3$ and so must map more than $3$ integers into images distinct from their pre-images. In other words, $f$ has a representation of the form

$$f = (123 ...) ( ... ) ... .$$

Moreover, $f$ can't be of the form $(1,2,3,m)$ since this is an odd permutation (WHY?) Thus $f$ must map at least two integers $>3$, say $4$ and $5$, into elements distinct from $4$ and $5$, respectively. Now let $g = (3,4,5)\in A_n$. Then $h = gfg^{-1} \in N$, but $h = gfg^{-1} = g (1,2,3, ...)g^{-1} g ( ... )g^{-1} ... g^{-1}$ (WHY?), and so $h = (1,2,4 ...) (... ) ...$. Now if $j > 5$ and $f(j) = j$, then clearly $h(j) = j$ (since $h = gfg^{-1}$), so However, $f^{-1}h \in N$, and $f^{-1}h(1) = 1$. In other words, $f^{-1}h \in N$, $f^{-1}h \not= (1)$ (WHY?), and $f^{-1}h$ leaves fixed more elements than $f$. This contradicts the choice of $f$. Thus this case is eliminated.

Case 2

$f$, in its representation as a product of disjoint cycles, contains at least $2$ distinct transpositions, i.e., $f$ is of the form

$$f = (12) (34) ...$$

Again, choose $g = (3,4,5)\in A_n$. Then as above

$$h = gfg^{-1} = (1,2) (4,5) ... .$$

As before $f^{-1}h \in N$, and $f^{-1}h(j) = j$ if $f(j) = j$ and $j > 5$. It is possible though here that $f$ leaves $5$ fixed, whereas $f^{-1}h(5) = 4$. However $f^{-1}h(1) = 1$ and $f^{-1}h(2) = 2$. Thus again $f^{-1}h \not= (1)$ (WHY?) and has more fixed points than $f$, a contradiction.

Since the two cases exhaust the possible representations for $f$ other than $f$ being a 3-cycle, we conclude that $f$ must indeed be a 3-cycle, and by the first part of the proof, we than get $N =A_n$. $\Box$

On the basis of this theorem and exercise 4 of Section 6.1, we can easily get further examples which show the converse of Lagrange's theorem is false. For example, $A_5 = 5!/2 =60$, but $A_5$ has no subgroup of order $30$ for such a subgroup would be normal, whereas we know $A_5$ is simple.

We conclude this section and chapter with a nice application of factor groups and of some of our earlier results (i.e., Theorem 6.3.4 obtained from Proposition 6.3.3). With this in mind, we first prove the following result.

Proposition 6.3.3   : If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof: We will write $Z = Z(G)$, the center of $G$. We first remark that $G/Z$ is defined since $Z \lhd G$ (WHY?). Now since $G/Z$ is a cyclic group, let's write $G/Z = \langle aZ\rangle$ for $a \in G$. Since

$$G=\cup_{n\in \mathbb{Z}} a^n Z,$$

if $g$ and $h$ are any elements of $G$ then $g \in a^k Z$ and $h \in a^m Z$ for integers $k,m$. Thus $g = a^kz_1$ and $h = a^mz_2$ where $z_1,z_2 \in Z$. Then

$$gh=(a^kz_1)(a^mz_2)=a^ka^mz_1z_2=(a^mz_2)(a^kz_1)=hg.$$

Hence $G$ is abelian. $\Box$

We can now obtain our desired application, i.e.,

Theorem 6.3.4   A group of order $p^2$ is abelian.

Proof: Let $G$ be a group such that $\vert G\vert = p^2$ and let $Z = Z(G)$ be the center of $G$. By Theorem 4.3.5, we know that $Z$ is non-trivial. Thus from Lagrange's Theorem $\vert Z\vert = p$ or $=p^2$. If $\vert Z\vert = p^2$, then $G=Z= Z(G)$, and so $G$ is abelian. On the other hand if $\vert Z\vert = p$, then we consider $G/Z$. Now $\vert G/Z\vert = p^2/p = p$. Thus by exercise 4 of Section 5.1, $G/Z$ is cyclic. Hence $G$ is again abelian by Proposition 6.3.3, and actually $\vert Z\vert = p^2$. $\Box$

Actually one can prove more and also show that there are precisely two non-isomorphic groups of order $p^2$, but we shall not go into these enumeration matters here.