# Show that a group with 21 elements contains at max 3 subgroups with 7 elements

I don't know if lagrange's theorem apply here anyway. I just know that this group can have a subgroup with 3 elements, but I don't know about any theorem that talks about number of elements of the subgroups.

I've searched and found something about sylow theorems, but I don't think I'm allowed to use that. How do I proceed?

Thank you so much!

• Hint: count it. Suppose there are 4 such subgroups. What is the intersection of two of them? And it is off the topic, but let you know that Sylow's theorem can be used to find exactly how many such subgroups are there. In your casex, there is only one such subgroup. – Rubertos Jul 7 '15 at 22:51
• Note (in keeping with Rubertos's hint) that these 7 element subgroups are cyclic, i.e. any element other than the identity will generate the full 7 element subgroup. – hardmath Jul 7 '15 at 22:56

Let $G$ be a group of order $21$. Any subgroup of order $7$ will be cyclic, and is generated by any non-trivial element of the group, so if $H,K$ are subgroups of $G$ of order $7$, and $H \cap K\neq \{id\}$, then $H=K$. In other words, the intersection of any two arbitrary subgroups of order $7$ must be trivial. So if $G$ has at least $4$ distinct subgroups of order $7$, then $G$ must have at least $24$ elements not equal to the identity element, $id$. But $G$ has order $21$, a contradiction. $\square$