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

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

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.

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

• What is the obvious subgroup of order $17$ in $H_{7}H_{17}$? – BAYMAX Feb 18 '17 at 13:35
• A cyclic group of order $7\cdot 17 = 119$ has an element $g$ of order $119$. $g^7$ has order $17$, and thus generates a (cyclic) subgroup of order $17$. That's the obvious subgroup of order $17$. Alternatively, the identity element along with the $16$ elements of order $17$ (which will turn out to be $g^7, g^{14}, g^{21},$ and so on). – Arthur Feb 18 '17 at 13:38
• Wow it's nice, now how is $g^{7}$ normal? – BAYMAX Feb 18 '17 at 13:41
• oh is that $g^{7}$ is a subgroup of a normal subgroup here $H_{7}H_{17}$ ,and hence normal? – BAYMAX Feb 18 '17 at 13:44
• Because the identity element has order $1$, not $17$, and because it's the only element where the different order-$17$-groups intersects. That means that for each of the $35$ order-$17$-groups, there are $16$ elements that it has to itself. That makes it $35\cdot 16$ elements of order $17$. – Arthur Feb 18 '17 at 14:28

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.