TITLE: Elementary question in Group Theory with less prerequisite
QUESTION [7 upvotes]: Here I am posing a problem, which my beginning students of algebra were discussing for long time.
Question: Without using theorem of Cauchy or Sylow, can we show that a group of order $15$ contains elements of order $3$ and $5$?
One can use Lagrange's theorem here. 

I know the solution using Cauchy theorem. But these things I have not yet taught in the class.

REPLY [2 votes]: Given two distinct subgroups of order $5$ $H$ and $K$, $H \cap K = \{e\}$ (because otherwise the element in common would generate them both).
So if all elements except $1$ have order $5$, the number of elements in the group must be $4n + 1$, which $15$ isn't.
So that means that there must be an element of order $3$.
Now assume that all elements have order 3. Again, by the same reasoning the number of elements in the group must be in the form $2n + 1$. So there are $n = 7$ subgroups of order $3$. I can't get further than that.