1
$\begingroup$

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.

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

$\endgroup$
7
  • $\begingroup$ What is the obvious subgroup of order $17$ in $H_{7}H_{17}$? $\endgroup$ – BAYMAX Feb 18 '17 at 13:35
  • 2
    $\begingroup$ 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). $\endgroup$ – Arthur Feb 18 '17 at 13:38
  • $\begingroup$ Wow it's nice, now how is $g^{7}$ normal? $\endgroup$ – BAYMAX Feb 18 '17 at 13:41
  • $\begingroup$ oh is that $g^{7}$ is a subgroup of a normal subgroup here $H_{7}H_{17}$ ,and hence normal? $\endgroup$ – BAYMAX Feb 18 '17 at 13:44
  • 1
    $\begingroup$ 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 $17$. $\endgroup$ – Arthur Feb 18 '17 at 14:28
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.