There is an excercise in Rotman's Introduction to the theory of groups: which of the following properties, when enjoyed by both $K$ and $Q$, is also enjoyed by every extension of $K$ by $Q$?

  1. finite
  2. $p$-group
  3. abelian
  4. cyclic
  5. solvable
  6. nilpotent
  7. ACC (ascending chain condition)
  8. DCC (descending chain condition)
  9. periodic (every element has finite order)
  10. torsion-free (every element other than 1 has infinite order).

I would say that the answer is at least being finite, $p$-group, solvable: if $K, Q$ are finite, then every extension $G$ of $K$ by $Q$ has order $|K| \cdot |Q|$. Solvability follows from another theorem, if $H$ and $G/H$ are solvable, then so is $G$.

The symmetric group $S_3$ is an extension of $\mathbb Z_3$ by $\mathbb Z_2$, hence 3. and 4. (being abelian or cyclic) aren't closed under group extension.

  • 2
    $\begingroup$ The symmetric group $S_3$ is also a counter-example for 6 (its center is trivial). $\endgroup$ – Clément Guérin Jul 12 '16 at 10:43

[7.] Not sure what you mean here. Are you talking about Noetherian groups? If so, then you can show that any subgroup $H$ of $G$ is finitely generated, given that $K$ and $G/K$ are Noetherian for some $K\trianglelefteq G$. This follows from the facts that $H\cap K\leq K$ is finitely generated and that $H/(H\cap K)\cong HK/K\leq G/K$ is also finitely generated.

[8.] Not sure what you mean here either. Are you talking about Artinian groups? If so, the answer is also yes. Consider the chain of normal subgroups $G=G_0\geq G_1\geq G_2\geq \ldots$. Suppose that a normal subgroup $K$ of $G$ satisfies $K$ and $G/K$ being Artinian. The image of this chain under $G\to G/K$ is given by $G/K=H_0\geq H_1\geq H_2\geq \ldots$. Then, by the assumption that $G/K$ is Artinian, we get $H_q=H_{q+1}=H_{q+2}=\ldots=:\bar{H}$ for some $q\in\mathbb{N}_0$. Now, we look at the sequence of subgroups of $K$, $K=K_0\geq K_1\geq K_2\geq \ldots$, where $K_i:=K\cap G_i$ for $i=0,1,2,\ldots$. Note that, for some integers $k\geq q$, $K_k=K_{k+1}=K_{k+2}=\ldots=:\bar{K}$. Thus, for $l=k,k+1,k+2,\ldots$, we have the exact squence $1 \to \bar{K} \to G_l \to \bar{H}\to 1$. By the Five Lemma for groups, $G_k=G_{k+1}=G_{k+2}=\ldots$.

[9.] If $1\to K \to G \to Q\to 1$ is exact with $K$ and $Q$ being periodic, then every $x\in G$ satisfies $x^q\mapsto 1_Q$ via $G\to Q$ for some integer $q>0$. Thus, $x^q$ is in the image of $K\to G$. As $K$ is periodic, $x^{qk}=\left(x^q\right)^k=1_G$ for some integer $k>0$.

[10.] If $1\to K\to G\to Q\to 1$ is exact with $K$ and $Q$ being torsion-free, then $x\in G$ with $x^q=1_G$ for some integer $q>0$ implies $x^q\mapsto 1_Q$ via $G\to Q$. As $Q$ is torsion-free, $x\mapsto 1_Q$ via $G\to Q$. Hence, $x$ is in the image of $K\to G$. Since $K$ is torsion-free and the map $K\to G$ is injective, $x^q=1_G$ happens only when $x=1_G$.

  • $\begingroup$ The problem with proving these properties came from my misconception that every group extensions is realised by a semidirect product which is just dead wrong. +1. $\endgroup$ – Santiago Jul 12 '16 at 16:24

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