I have already found two definitions for a Baer group.
$G$ is a Baer group if it is generated by all cyclic subnormal subgroups.
$G$ is a Baer group if every cyclic subgroup is subnormal.
I want to prove the equivalence of the two definitions. Obviously, (2) implies (1). Please help me with the converse.
Claim. If group is generated by cyclic subnormal groups, then every finitely generated subgroup is subnormal.
Proof. Let $A, B < G$ be f. g., nilpotent, and subnormal. We want to prove that $C := \langle A, B \rangle$ has this properties. (Then every cyclic will be subnormal, because every subgroup of nilpotent group is subnormal — it's easy).
First, note that $C$ lies in Hirsch radical — more or less by definition, therefore (locally, but this doesn't matter because it is f. g.) nilpotent.
Now you want to prove that $C$ is subnormal. I'll sketch the proof, it's pretty straightforward.
Let's look at some subnormal chain from $A$ to $G$, $A = U_0 \lhd \dots \lhd U_k = G$. First, replace $A$ by $A' = A^C$, so that $A' \lhd C$. Look at partial subnormal chain $A^{U_i}$. Take $B$-invariant subgroups of them; it's also subnormal chain, say, $V_i$, going from $A$ to $A^G$. Finally observe that $V_i \lhd V_{i+1} B$, $B$ subnormal in $V_{i+1} K$. It immediately gives that $V_i B$ is subnormal in $V_{i+1} B$, and because $A^G B$ is subnormal in $G$ we're done.
I guess it's written in Baer's articles near late 40s somewhere, but I do not remember particular reference.
