Elementary question in Group Theory with less prerequisite 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.
 A: We assume the group is not cyclic. If it generates $G$ we are done. Otherwise by Lagrange it has order $3$ or $5$. If the order is $5$ then then let $C$ be the group generated by $g$. Since the index of $C$ is $3$ and $3$ is the smallest prime dividing $|G|$ we conclude that $C$ is normal and so we take $G\rightarrow \frac{G}{C}$ the natural isomorphism $\varphi$. We take a generator of $\frac{G}{C}$ and consider its preimage. call this preimage $h$ and the subgroup generated by it $H$. Since $H$ maps to all of $\frac{G}{C}$ we apply the first isomorphism theorem to the restriction of the natural isomorphism to $H$ and so $\frac{H}{ker(\varphi|_H)}\cong\frac{G}{C}$. This tells us the order of $H$ is a multiple of $3$ and since $H$ is not cyclic this means the order of $H$ is $3$.
So if there is an element of order $5$ there is an element of order $3$.
I'm not sure how to prove it the other way around.
A: 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.
