Be $G$ a group of order $35,$ Prove at least $1$ Element lies not in a subgroup of order $7$

Asked Oct 04 '17 at 14:21

Active Oct 04 '17 at 14:44

Viewed 69 times

I am an exchange student here and have difficulties solving the first question of my group theory course tomorrow.

The Question is: Let $G$ be a group of order $35.$ Prove that at least one element of $G$ does not lie in any subgroup of order $7.$ Hint: If $H$ and $K$ are distinct subgroups of $G$ such that both have order $7,$ what can you say about the order of the intersection of $H$ and $K?$

The only theorem we have learned is Lagranges Theorem, but I am not sure where it can help me. I have no idea how I should prove this, so please help me:(

edited Oct 04 '17 at 14:44

JohnColtraneisJC

1,919

asked Oct 04 '17 at 14:21

TwentyTwo

Do you know Cauchy's theorem? – Bernard Oct 04 '17 at 14:24
Hint : $G$ must be cyclic and has an element with order $5$ – Peter Oct 04 '17 at 14:30
never heard of Cauchys theorem, so I think I am not allowed to use it:/ – TwentyTwo Oct 04 '17 at 14:44

Be $G$ a group of order $35,$ Prove at least $1$ Element lies not in a subgroup of order $7$

0 Answers0