Prove that if a group $G$ is generated by all cyclic subnormal subgroups, then every cyclic subgroup is subnormal I have already found two definitions for a Baer group. 


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*$G$ 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.
 A: 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.
