Math Problems Vault

Problem III.2.1.1

A chain of subgroups

e=G_{0} \subset G_{1} \subset \ldots \subset G_{n} \subset G_{n+1}=G

where $G_{i-1} \triangleleft G_{i}, 1 \leqslant i \leqslant n+1$ , is called a normal series of the group $G$ . A series of decreasing subgroups is also called normal

G=G_{0} \supset G_{1} \supset \ldots \supset G_{n} \supset G_{n+1}=e

where $G_{i+1} \triangleleft G_{i}$ .

If all the terms of series (*) are distinct and $G_{i-1} \triangleleft H \triangleleft G_{i} \Longrightarrow H=G_{i-1}$ or $H=G_{i}$ for all $i$ , one speaks of a composition series of the group $G$ . The quotient groups $F_{i}=G_{i} / G_{i-1}$ are in this case called composition factors.

Show that:

any normal series of a finite group can be refined to a composition series by inserting additional terms into it as long as this is possible;
the factors $F_{i}$ are simple groups (or cyclic groups of prime order);
the group $G$ is solvable exactly when all its composition factors are cyclic groups of prime order. Let us mention without proof the Jordan--Hölder theorem, according to which the set of composition factors of the group $G$ , up to isomorphism and order of listing, does not depend on the choice of composition series.

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Problem III.2.1.2

Prove that any finite $p$ -group is solvable.

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Problem III.2.1.3

(a)(b)(c)

Prove the simplicity of the alternating group $A_{n}, n \geqslant 5$ , following the scheme of argument outlined below.

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(a)

In the normal subgroup $K \triangleleft A_{n}, K \neq e$ , one should take a permutation $\pi \neq e$ fixing the maximum possible number $m$ of symbols from $\Omega =\left\{ 1,2, \ldots , n\right\}$ . If $m=n-3$ , then $\pi =(i j k)$ and $K=A_{n}$ (see [ВА I, Ch. $4, \S 2$ , Exercise 11]), so we assume $m<n-3$ .

(b)

If $\pi =(123 \ldots ) \ldots$ is the decomposition of $\pi$ into independent cycles, then the parity of $\pi$ and the condition $m<n-3$ imply the inequality $m<n-5$ . It is also possible that $\pi =(12)(34) \ldots$ consists of independent cycles of length 2.

(c)

In any case, consider the commutator $(\pi , \sigma )=\pi \sigma \pi^{-1} \sigma^{-1} \neq e$ with $\sigma =(345)$ , and verify that it fixes more than $m$ symbols. This contradicts the choice of $m$ and proves the statement.

Problem III.2.1.4

Show that the alternating group $A_{5}$ contains no subgroups of orders 15 and 20.

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