## Equivalent characterizations of solvability for groups

Let $G$ be a finite group. Prove that the following are equivalent.

1. $G$ is solvable.
2. $G$ has a subnormal series $1 = H_1 \vartriangleleft \cdots \vartriangleleft H_k = G$ such that $H_{i+1}/H_i$ is cyclic for $1 \leq i < k$.
3. All composition factors in any composition series of $G$ have prime order.
4. $G$ has a subnormal series $1 = H_1 \vartriangleleft \cdots \vartriangleleft H_k = G$ such that $H_i$ is normal in $G$ and $H_{i+1}/H_i$ is abelian for $1 \leq i < k$.

5. We begin with some lemmas.

Lemma 1: Let $G$ be a finite group and $N \leq G$ a normal subgroup such that $G/N$ is abelian. Then there exists a chain of subgroups $N = K_1 \vartriangleleft K_2 \vartriangleleft \cdots \vartriangleleft K_\ell = G$ such that $K_{i+1}/K_i$ has prime order for all $1 \leq i < \ell$. Proof: $G/N$ is a finite group, so by a previous exercise there exists a composition series $1/N = K_1/N \vartriangleleft K_2/N \vartriangleleft \cdots \vartriangleleft K_\ell/N = G/N$ such that $(K_{i+1}/N)/(K_i/N)$ is simple for all $1 \leq i < \ell$. Now because $G/N$ is abelian, each $K_i/N$ is abelian. We saw in a previous exercise that the only abelian simple groups are (up to isomorphism) $\mathbb{Z}/(p)$ for primes $p$, each $(K_{i+1}/N)/(K_i/N)$ has prime order. Using the Lattice Isomorphism Theorem we have a chain of subgroups $N = K_1 \vartriangleleft \cdots \vartriangleleft K_\ell = G$ and for all $1 \leq i < \ell$, $K_{i+1}/K_i \cong (K_{i+1}/N)/(K_i/N) \cong \mathbb{Z}/(p)$ by the Third Isomorphism Theorem. $\square$

Lemma 2: If $G$ is simple and solvable, then $G \cong \mathbb{Z}/(p)$ for some prime $p$. Proof: If $G$ is simple, then the only subnormal series of $G$ is $1 \vartriangleleft G$. Then $G$ is abelian; we saw previously that every abelian simple group is isomorphic to $\mathbb{Z}/(p)$ for some prime $p$. $\square$

Lemma 3: If $1 = H_0 \vartriangleleft H_1 \vartriangleleft \cdots \vartriangleleft H_k = G$ is a composition series where $G$ is solvable and finite, then $H_{i+1}/H_i$ is abelian for all $0 \leq i < k$. Proof: We proceed by induction on $k$. If $k = 0$, then $G = 1$ and the conclusion that $H_{i+1}/H_i$ is abelian for all $0 \leq i < 0$ holds vacuously. Suppose now that for all composition series of length $n$ such that $G = H_n$ is solvable we have all composition factors abelian, and let $1 = H_0 \vartriangleleft \cdots \vartriangleleft H_n \vartriangleleft H_{n+1} = G$ be a composition series such that $G$ is solvable. Now $H_n \leq G$ is solvable by a previous exercise, and $1 = H_0 \vartriangleleft \cdots \vartriangleleft H_n$ is a composition series for $H_n$ of length $n$. By the induction hypothesis, we have $H_{i+1}/H_i$ abelian for all $0 \leq i < n$. Moreover, $H_{n+1}/H_n$ is simple and solvable by a previous exercise, so that $H_{n+1}/H_n \cong \mathbb{Z}/(p)$ is abelian. By induction the conclusion holds for all composition series which terminate with a solvable group. $\square$

Lemma 4: Let $G$ be a group and $H \leq G$ a subgroup. Then $\bigcap_{g \in G} gHg^{-1}$ is a normal subgroup of $G$. Proof: Suppose $x,y \in \bigcap_{g \in G} gHg^{-1}$. Then for all $g \in G$, we have $x = gag^{-1}$ and $y = gbg^{-1}$ for some $a,b \in H$. Now $(gag^{-1})(gb^{-1}g^{-1}) = g(ab^{-1})g^{-1} \in gHg^{-1}$; by the Subgroup criterion, $\bigcap_{g \in G} gHg^{-1}$ is a subgroup of $G$. Now let $h \in G$; then $h(\bigcap_{g \in G} gHg^{-1})h^{-1} = \bigcap_{g \in G} hgHg^{-1}h^{-1} = \bigcap_{k^\prime \in G} kHk^{-1}$; hence this subgroup is normal in $G$. $\square$

Lemma 5: Let $G$ be a finite nonsimple group such that every composition factor of every composition series of $G$ has prime order and let $M \leq G$ be a minimal nontrivial normal subgroup. Then $M$ is abelian. Proof: Let $M$ be such a subgroup of $G$. By a previous exercise, there exists a composition series of $G$ which contains $M$ as a term, say $1 = K_0 \vartriangleleft K_1 \vartriangleleft \cdots \vartriangleleft K_\alpha = M \vartriangleleft \cdots \vartriangleleft K_\ell = G$. Let $N = K_{\alpha - 1}$; by our hypothesis, $H_\alpha/H_{\alpha - 1}$ has prime order $p$ and thus is isomorphic to the abelian simple group $\mathbb{Z}/(p)$. Thus for all $x,y \in M$, we have $(xy)N = (yx)N$, hence $x^{-1}y^{-1}xy \in N$. Now let $x,y \in M$ and $g^{-1} \in G$ be arbitrary. Because $M$ is normal in $G$, we have $g^{-1}xg = \overline{x} \in M$ and $g^{-1}yg = \overline{y} \in M$. By the previous point, $\overline{x}^{-1}\overline{y}^{-1}\overline{x}\overline{y} \in N$. Thus $g^{-1}x^{-1}gg^{-1}y^{-1}gg^{-1}xgg^{-1}yg = g^{-1}x^{-1}y^{-1}xyg \in N$, hence $x^{-1}y^{-1}xy \in gNg^{-1}$. Now by Lemma 4, $\bigcap_{g \in G} gNg^{-1} < M$ is normal in $G$. Because $M$ is a minimal nontrivial normal subgroup of $G$ and $N$ is strictly contained in $M$, we have $\bigcap_{g \in G} gNg^{-1} = 1$, hence $x^{-1}y^{-1}xy = 1$, hence $xy = yx$. Thus $M$ is abelian. $\square$

Now to the main result. We prove the following chains of implications: $(1) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$ and $(1) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$.

Note that $(3) \Rightarrow (2)$ because every group of prime order is cyclic and $(2) \Rightarrow (1)$ because every cyclic group is abelian. $(4) \Rightarrow (1)$ is trivial.

$(1) \Rightarrow (3)$ Let $G$ be solvable and let $1 = H_1 \vartriangleleft \cdots \vartriangleleft H_k = G$ be an arbitrary composition series of $G$. By Lemma 3, all composition factors are abelian simple groups. We saw in a previous exercise that every such group is isomorphic to $\mathbb{Z}/(p)$ for some prime $p$, thus all composition factors have prime order.

$(3) \Rightarrow (4)$ Let $G$ be a finite group with the property that every composition factor of every composition series has prime order.

If $|G| = 1$, then $G = 1$ and it is vacuously true that $G$ has a normal series with all intermediate quotients abelian.

Suppose that every finite group $H$ with $|H| \leq n$ has a normal series with all intermediate quotients abelian, and supose $G$ is a group of order $n+1$.

If $G$ is simple, then $G$ has only one composition series, namely $1 \vartriangleleft G$, and so $G$ has prime order by the hypothesis. Thus $G \cong \mathbb{Z}/(p)$ for some prime $p$ and $G$ is abelian; so a normal series with all intermediate quotients abelian exists.

If $G$ is not simple, then the set $A$ of nontrivial normal subgroups of $G$ is not empty. Now $A$ contains a minimal element $M$ which, by Lemma 5, is abelian. By the induction hypothesis, since $G/M$ is a finite group of order less than or equal to $n$, we have a normal series $1 = H_0/M \vartriangleleft \cdots \vartriangleleft H_k/M = G/M$ where $(H_{i+1}/M)/(H_i/M)$ is abelian. By the Third Isomorphism Theorem, $H_{i+1}/H_i$ is abelian, and by the Lattice Isomorphism Theorem, $H_i \leq G$ is normal. Thus $1 \vartriangleleft M = H_0 \vartriangleleft \cdots \vartriangleleft H_k = G$ is a normal series for $G$ with all intermediate quotients abelian.

Another approach to 1=>3: Let $G$ be solvable. It has a subnormal series with abelian factor groups. By Schreier refinement theorem, it has a refinement to a composition series, The factor groups of this series are abelian since they are quotients of abelian groups. So these are simple abelian, thus have prime orders by a previous exercise(3.4#1).