A group of order $595$ has a normal subgroup of order $17$

Question

I was referring this question, which i understood but the last three lines of Don Antonio's answer -

A group of order $595$ has a normal Sylow 17-subgroup..

Any help with the last three lines-

"But then we're done since $H_7H_{17}$ is a cyclic group with an obviously normal subgroup of order $\;17\;$, and normal subgroup of normal cyclic subgroup is normal itself, i.e.

$$A\lhd B\lhd G \text{ and }B\text{ cyclic}\implies A\lhd G."$$

Answer 1

In general you can apply two facts: every subgroup $A$ of a cyclic group $B$ is characteristic (that is, being fixed by any automorphism of $B$, and we write $A$ char $B$). Secondly, if $A$ char $B \unlhd G$, then $A \unlhd G$. The proofs of these two statements are straightforward and by the way can be found on this site, see here or here.