Series of groups; solvable groups revisited

We shall apply the third isomorphism theorem presently, but first we introduce a special sequence of subgroups of a group (usually called series of groups). This was alluded to in the introduction to this chapter. Let

\begin{displaymath} \{e\} =G_{t+1}\subset G_t\subset ...\subset G_2 \subset G_1=G\end{displaymath} (11.7)

be a sequence of subgroups of the group $G$ where each $G_{i+1}$ is normal in $G_i$ (but not necessarily in all of $G$). Such a sequence of subgroups is called a normal series for $G$. Associated with a normal series for a group $G$, is an associated sequence of factors (or factor groups); viz

\begin{displaymath} G_1/G_2, G_2/G_3, ..., G_t/G_{t+1} = G_t. \end{displaymath}

We observe that a normal series always exists for an arbitrary group $G$. We could, e.g., take the trivial normal series: $\{e\}\subset G = G_1$. There is also nothing unique about a normal series, e.g., the symmetric group $S_4$ has the following normal series, among others:

\begin{displaymath} \begin{array}{c} \{e\} \subset C_2\subset V_4\subset A_4\sub... ...e\} \subset V_4\subset S_4, \ \{e\} \subset S_4, \end{array}\end{displaymath} (11.8)

where $V_4$ is the Klein $4$-group (see Section 6.3), and, as usual, $C_n$ denotes a cyclic group of order $n$, here, e.g., take $C_2 = \langle (12)(34)\rangle$. Note that all terms in the fifth series given above for $S_4$ occur in the fourth and all those in the fourth occur in the third and those in the third occur in the first. A similar situation prevails between the fifth, fourth, second, and first series. This illustrates the following: one normal series is called a refinement of another if all the terms of the second occur in the first series. Hence the second series above is a refinement of the fourth series. The third series is also a refinement of the fourth series. However, the second series is not a refinement of the third series.

Finally, two normal series

\begin{displaymath} \begin{array}{c} \{e\} =G_{s+1}\subset G_s\subset ...\subset... ...+1}\subset H_t\subset ...\subset H_2 \subset H_1=H, \end{array}\end{displaymath} (11.9)

are called equivalent or (isomorphic) if there exists a 1-1 correspondence between the factors of the two series (thus $s = t$) such that the corresponding factors are isomorphic.

Example 11.2.1   Consider the two series for $\mathbb{Z}_{15}$,

\begin{displaymath} \begin{array}{c} \mathbb{Z}_{15} \supset \langle [5]\rangle... ...Z}_{15} \supset \langle [3]\rangle \supset \{[0]\}, \end{array}\end{displaymath}

where $[5]$ denotes the residue class of $5$ mod $15$. These normal series are equivalent: For $\mathbb{Z}_{15}/\langle [5]\rangle \cong \mathbb{Z}_5$, $\langle [5]\rangle /\{[0]\} \cong \mathbb{Z}_3$, while the factors of the second series are $\mathbb{Z}_{15}/\langle [3]\rangle \cong \mathbb{Z}_3$, $\langle [3]\rangle /\{[0]\} \cong \mathbb{Z}_5$ (where $[3]$ denotes the residue class of $3$ mod $15$).

Our first general theorem in this spirit is the following important theorem due to Schreier.

Theorem 11.2.2 (Schreier)   Any two normal series for a group G have equivalent refinements.

Proof: Consider two normal series for G as in (11.9). Define

\begin{displaymath} \begin{array}{c} G_{ij} = (G_i \cap H_j)G_{i+1}, \ \ \ \ j =... ..._i \cap H_j) H_{j+1}, \ \ \ \ i = 1, 2, ..., s + 1. \end{array}\end{displaymath}

Then we have

\begin{displaymath} \begin{array}{c} G = G_{11}\supset G_{12}\supset ... \supse... ...,s+1} = G_3 \supset ... \supset G_{t,s+1} = \{e\}, \end{array}\end{displaymath}

and

\begin{displaymath} \begin{array}{c} G = H_{11}\subset H_{12}\supset ... \supse... ...,t+1} = H_3 \supset ... \supset H_{s,t+1} = \{e\}, \end{array}\end{displaymath}

Now applying the third isomorphism theorem (Theorem 11.1.1) to the groups $G_i$, $H_j$, $G_{i+1}$, $H_{j+1}$, we have that $G_{i,j+1}= (G_i \cap H_{j+1})G_{i+1} \lhd G_{i,j} = (G_i \cap H_j)G_{i+1}$ and $H_{j,i+1} = (G_{i+1}\cap H_j)H_{j+1} \lhd H_{j,i} = (G_i \cap H_j)H_{j+1}$. Furthermore, also by Theorem 11.1.1,

\begin{displaymath} G_{ij}/G_{i,j+1}\cong H_{ji}/H_{j,i+1}. \end{displaymath}

Thus the above two are normal series which are refinements of the two given series and they are equivalent. $\Box$

We shall, in the next section, apply Schreier's Theorem 11.2.2 to obtain the important theorem of Jordan-Hölder, but first we wish to give an alternate characterization of solvable groups. Before doing this, we establish the following useful theorem.

Theorem 11.2.3   If $\{e\} =G_{t+1}\subset G_t\subset ...\subset G_2 \subset G_1=G$ is a normal series for the group $G$ and if $H \leq G$, then

\begin{displaymath} H = H \cap G_1 \supset H \cap G_2\supset ... \supset H \cap G_{t+1} = {e} \end{displaymath}

is a normal series for $H$. The factors of the normal series in (11.2.3) are isomorphic to subgroups of the factors of the normal series for $G$.

Proof: We apply the second isomorphism theorem (Theorem 8.3.7) to the subgroups $G_{i+1}$ and $H \cap G_i$ of the group $G_i$. Hence, we have


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First note that since $G_{i+1}\lhd G_i$, $G_{i+1}(H \cap G_i)$ is a subgroup of $G_i$, Proposition 8.3.6 implies $G_{i+1}\lhd G_{i+1}(H \cap G_i)$, and also $H\cap G_{i+1}\lhd H \cap G_i$. Second note that $G_{i+1}\cap (H \cap G_i) = H \cap G_{i+1}$ since $G_{i+1}\subset G_i$ and that is how we get the group at the lower vertex of our diagram. Thus Theorem 8.3.7 yields that

\begin{displaymath} H \cap G_{i}/H \cap G_{i+1} \cong G_{i+1} (H \cap G_{i})/G_{i+1} \cong G_i/G_{i+1}. \end{displaymath}

$\Box$

We now make the following definition (cf. Definition 8.1.6).

Definition 11.2.4   A group $G$ is said to be solvable if it has a normal series all of whose factors are abelian groups.

Since we have already defined a solvable group in Section 8.1, we must show that these two definitions are equivalent. Thus suppose that $G$ is solvable according to Definition 8.1.6. Then

\begin{displaymath} G\supset G'\supset G''\supset ... \supset G^{(t)}= \{e\}, \end{displaymath}

where the superscripts designate the higher commutator subgroups (see Section 8.1). This is, of course, a normal series (see comments at the beginning of Section 8.1 and Theorem 8.1.5) for $G$. Moreover, $G^{(i)}/G^{(i+1)}$ is abelian by Theorem 8.1.4. Hence $G$ is solvable according to Definition 11.2.4 above.

Suppose now that G is solvable in the sense of Defintion 11.2.4. So that G has a normal series as in (11.7), such that each factor $G_i/G_{i+1}$ is abelian. In particular, $G/G_2 = G_1/G_2$ is abelian. Thus by Theorem 8.1.4, $G_2\supset G_1'=G'$. Since $G_2/G_3$ is abelian, we have, again by Theorem 8.1.4, that $G_3\supset G_2'\supset (G_1')'=G''$. Similarly,

\begin{displaymath} G_4\supset G_3'\supset (G_2')'=G_1'''=G'''. \end{displaymath}

Continuing in this fashion, we finally get that

\begin{displaymath} \{e\} = G_{s+1} \supset G^{(s)}. \end{displaymath}

Hence $G^{(s)} = \{e\}$, and $G$ is solvable according to our original definition.

Thus we are at liberty to use whichever characterization of solvabilty is more convenient. In the following theorem, we arbitrarily use the characterization of solvability introduced in this section. We strongly advise the reader to prove the theorem (see exercise 3 for this section) using the initial definition (Definition 8.1.6) without making use of the equivalent characterization we have just established.

Theorem 11.2.5   Any subgroup and any factor group of a solvable group is solvable.

Proof: Suppose $G$ is solvable. Then $G$ has a normal series (11.7), such that $G_i/G_{i+1}$ is abelian. Let $H \leq G$. Then form the series (11.2.3). By Theorem 11.2.3, we thus get a normal series, and $(H \cap G_i)/(H \cap G_{i+1})$ is isomorphic to a subgroup of $G_i/G_{i+1}$ and is, therefore, abelian. This completes the first part of the theorem.

Again let $G$ be a solvable group and let (11.7) again denote a normal series for $G$ with abelian factors. It is easy to see that any refinement of the series (11.7) also has abelian factors; e.g., suppose

\begin{displaymath} G_1\supset G_2\supset H\supset G_3\supset .... \end{displaymath}

is a refinement of (11.7). Then $H/G_3\subset G_2/G_3$ and, hence, is abelian. Since $G_2/H\cong (G_2/G_3)/(H/G_3)$, by Corollary 8.3.5, therefore, $G_2/H$ also abelian. Now let $N \lhd G$. Consider the normal series
\begin{displaymath} G \supset N \supset \{e\} . \end{displaymath} (11.10)

By Schreier's Theorem (Theroem 11.2.2), (11.7) and (11.10) have equivalent refinements. Let
\begin{displaymath} G \supset H_1\supset H_2 \supset ... \supset H_n\supset N\supset ...\supset \{e\} \end{displaymath} (11.11)

be a refinement of (11.10) equivalent to a refinement of (11.7). By our preceding observations, the factors of (11.11) are abelian. Since $(H_i/N)/(H_{i+1}/N)\cong (G_i/G_{i+1})$, by Corollary 8.3.5,

\begin{displaymath} G/N \supset H_1/N\supset H_2/N \supset ... \supset H_n/N\supset N/N= \{e\} \end{displaymath}

is a normal series for $G/N$ with abelian factors. $\Box$

Since $A_n$ for $n \geq 5$ was shown to be a simple group (see Theorem 6.3.2),

\begin{displaymath} A_n \supset \{e\} \end{displaymath}

is the only normal series for $A_n$, when $n \geq 5$. But $A_n$ for $n \geq 5$ is, of course, non-abelian, hence is not a solvable group. Consequently, by the preceding Theorem 11.2.5, $S_n$ for $n \geq 5$ is also not a solvable group.


Subsections
David Joyner 2007-08-06