1

Question:

Let $\left | G \right |=15$. If G has only one subgroup of order 3 and only one subgroup of order 5, prove that G is cyclic.

Looking for useful hints to the above question.

Thanks in advance.

Community
  • 1
Mathematicing
  • 6,253
  • Counterproof: If G were cyclic, it would be O. – Dronz Jun 24 '16 at 14:42
  • 2
    @Dronz What? $\Bbb Z/15\Bbb Z$ has that property. –  Jun 24 '16 at 14:56
  • 1
    Version without extra hypothesis about subgroups: http://math.stackexchange.com/questions/67407 Generalization to all orders with the hypothesis about subgroups: http://math.stackexchange.com/questions/1302635/a-finite-group-which-has-a-unique-subgroup-of-order-d-for-each-d-mid-n – Jonas Meyer Jun 24 '16 at 19:58
  • @G.Sassatelli It was a joke about the shape of the capital letter G. – Dronz Jun 24 '16 at 21:58

3 Answers3

12

Pick an element of $G$ not in either of those subgroups. What must that element's order be?

user2357112
  • 2,401
10

Some hinting:

(1) If a finite group has one unique subgroup of some given order, then that subgroup is normal

(2) If $\;N,H\lhd G\;$ and $\;G=NH\;$ , then in fact $\;G=NH\cong N\times H\;$

(3) Direct product of finite cyclic groups is cyclic if the groups' orders are coprime.

By the way, you don't need that "if" in the question: the condition is always fulfilled.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • Good point, 3 does not divide 4 :) – almagest Jun 24 '16 at 11:20
  • @almagest Thank you but I didn't get your point... – DonAntonio Jun 24 '16 at 11:22
  • I seem to remember that $p\ne1\bmod q$ and $q\ne1\bmod p$ is the standard test for groups of order $pq$. – almagest Jun 24 '16 at 11:23
  • @almagest Oh, that. Yes, of course you're right...but that is when you use Sylow theorems. In this case we aren't told it is possible...and besides we don't need it as it is assumed the subgroups are already normal. Thank you. – DonAntonio Jun 24 '16 at 11:40
  • "You don't need that "if""... tell that to Schur and Zassenhaus – MT_ Jun 24 '16 at 19:49
  • @Soke Thank you. I would, yet I think they're already pretty dead so I don't think they'll address my question...:) Anyway, my intention was to hint towards the fact that there's one unique group, up to isomorphism, of order $;15;$ : the cyclic one, and such a group always has normal Sylow subgroups. – DonAntonio Jun 24 '16 at 19:52
  • Yup. In general, if $p < q$ and $p \not\mid q-1$, then the only group of order $pq$ is the cyclic one. – MT_ Jun 24 '16 at 19:55
  • Could you please provide explanations on why these statements are true? – user569685 Aug 28 '20 at 22:26
  • @Hristmar Where are you stuck and what is your background in group theory? – DonAntonio Aug 28 '20 at 22:39
  • I dont even know where to begin proving those. My background is that i've failed my exam 2 times in a row. I've checked my notes but they statements are nowhere to be found. I assume they can be concluded from other statements but I'm still searching. – user569685 Aug 28 '20 at 23:13
  • @Hristmar Well, then perhaps you should write down a formal question for each thing that bothers you, and accompany it with your best shot at solving that problem that you can. – DonAntonio Aug 29 '20 at 07:05
3

Hint:

Let $T$ the subgroup of order $3$, $F$ the subgroup of order $5$. Show that each of them is normal in $G$ and $G=TF$. Then use the Chinese remainder theorem.

Bernard
  • 175,478