# Prove that $G$ is cyclic if $|G|=15$ and $G$ has only one subgroup each of orders $3$ and $5$

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.

• Counterproof: If G were cyclic, it would be O. Commented Jun 24, 2016 at 14:42
• @Dronz What? $\Bbb Z/15\Bbb Z$ has that property.
– user228113
Commented Jun 24, 2016 at 14:56
• Version without extra hypothesis about subgroups: math.stackexchange.com/questions/67407 Generalization to all orders with the hypothesis about subgroups: math.stackexchange.com/questions/1302635/… Commented Jun 24, 2016 at 19:58
• @G.Sassatelli It was a joke about the shape of the capital letter G. Commented Jun 24, 2016 at 21:58

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

– user228113
Commented Jun 24, 2016 at 17:06
• This is pure elegance! Commented Sep 19, 2018 at 9:07

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.

• Good point, 3 does not divide 4 :) Commented Jun 24, 2016 at 11:20
• @almagest Thank you but I didn't get your point... Commented Jun 24, 2016 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 $pq$. Commented Jun 24, 2016 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. Commented Jun 24, 2016 at 11:40
• "You don't need that "if""... tell that to Schur and Zassenhaus
– MT_
Commented Jun 24, 2016 at 19:49

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.