Group of order $105$ with a normal Sylow $3$-subgroup is cyclic

Asked Apr 19 '19 at 00:18

Active Apr 19 '19 at 00:21

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Let $G$ be a group of order $105 = 3 \times 5 \times 7$ with a normal Sylow $3$-subgroup, prove that $G$ is cyclic.

My attempt

I've been able to show that $G = NP$, where $N$ is the normal Sylow $3$-subgroup (which is of course cyclic) and $P$ is a cyclic group of order $35$. What's left to be shown is that $NP$ is cyclic. And I'm stuck.

Any help would be greatly appreciated. Thanks in advance!

edited Apr 19 '19 at 00:21

Shaun

44,997

asked Apr 19 '19 at 00:18

msd15213

3

Possible duplicate of Suppose $|G| = 105$. Show that it is abelian if it has a normal $3$-Sylow subgroup., since any finite abelian group of square-free order is cyclic. – Ege Erdil Apr 19 '19 at 07:52

Group of order $105$ with a normal Sylow $3$-subgroup is cyclic

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