I am reading John Beachy's online Abstract Algebra which has this question and its solution:
PROBLEM: Prove that if $G$ is a group of order 56, then $G$ has a normal Sylow 2-subgroup or a normal Sylow 7-subgroup.
SOLUTION: The number of Sylow 7-subgroups is either 1 or 8. Eight Sylow 7-subgroups would yield 48 elements of order 7, and so the remaining 8 elements would constitute the unique Sylow 2-subgroup.
I understand the first part of the solution, that if the number of $Syl_7(G) = s$, then $s = \{1, 8\}$, but I do have couple of questions for the rest of the paragraph, which I love to get explanation from you:
(1) He analyzed $s = 8$, why did he skip analyzing $s = 1$?
(2) I understand that $48 = 8 \times (7-1)$, 8 being the number of subgroups. How did he determine that each subgroup has 6 elements?
(3) When he wrote "... 48 elements of order 7," did he mean the smallest positive integer $m$ such that $g^m = e$, with $g$ being the group's element and $e$ the identity? Or did he mean the cardinality of the subgroup is 7?
(4) And finally, when he wrote "... and so the remaining 8 elements would constitute the unique ...," did this imply that sum of the order of all sylow subgroups will always equal to the order of the group, i.e. in this case, $|Syl_2(G)| + |Syl_7(G)| = |G|$?
Thank you very much for your time and effort.
POST SCRIPT (after feedback from "usermath" yesterday): ~~~~~~~~~~~~~~
I am very grateful to receive a great feedback from "usermath," and here are my comments on each of them. To other black-belt personalities other than "usermath," you are most welcome to pitch in:
(1) If G has 1 Sylow-7 subgroup then it is unique and it will be normal. So we are done.
[My comment: I got it now, thanks!]
(2) Cardinality of Sylow-7 subgroup is 7 and hence cyclic. So each of these has 6 elements of order 7 since identity has order 1.
[I am focusing on $|Syl_7(G)| = 7.$ My class-note specifically says that "Note that a subgroup $H$ of $G$ is a Sylow p-subgroup IF there exist integers $e$ and $m$ such that $|G| = p^em, (p,m) = 1,$ and $|H| = p^e.$" Judging from your response, don't you think it should be IFF instead? Keep in mind that my text is just a simple class-note, so it is not a commercial-grade text.]
(3) $g$ is element of order 7 means the smallest positive integer $m$ such that $g^m=e$ is 7.
[I got it now, thanks again!]
(4) Cardinality of any Sylow-2 subgroup is 8 and we are left with exactly 8 elemenents. And so these 8 elements form a Sylow-2 subgroup (Say $H_1$),since all other 48 elements has order 7 and 7 does not divide 8. Now if $H_2$ is another Sylow-2 subgroup different from $H_1$ then $H_2$ has a non identity element which is not in $H_1$. But all elements which are not in $H_1$ has order 7 and hence cannot be in $H_2$.
[I got all your points, but my question is actually a general one: Let's say that a group has more than one Sylow subgroups. Is there a theorem that says each of the group's element has to belong to the Sylow subgroups? In other words, is it possible to have an element that does not belong to any of the Sylow subgroup?]
And finally here is my own question: Don't you think that this question is badly worded from the beginning, since it is asking for proof that "$G$ has a normal Sylow 2-subgroup OR a normal Sylow 7-subgroup," but it ends up proving that "$G$ has a normal Sylow 2-subgroup AND a normal Sylow 7-subgroup"?
Thanks again to "usermath" and all others who have been reading it.
(1) If G has 1 Sylow-7 subgroup then it is unique and it will be normal. So we are done.
(2) Cardinality of Sylow-7 subgroup is 7 and hence cyclic. So each of these has 6 elements of order 7 since identity has order 1.
(3) $g$ is element of order 7 means the smallest positive integer $m$ such that $g^m=e$ is 7.
(4) Cardinality of any Sylow-2 subgroup is 8 and we are left with exactly 8 elemenents. And so these 8 elements form a Sylow-2 subgroup (Say $H_1$),since all other 48 elements has order 7 and 7 does not divide 8. Now if $H_2$ is another Sylow-2 subgroup different from $H_1$ then $H_2$ has a non identity element which is not in $H_1$. But all elements which are not in $H_1$ has order 7 and hence cannot be in $H_2$.