Group of order $15$ has element of order $5$ I need to show that group of order $15$ has element of order $5$. I need to do that without using Sylow theorems and their consequences and Cauchy theorem.
Basically i need to show that Group of order $15$ can't be consisted of $e$ and elements of order $3$.
 A: We only need to appeal to Lagrange's Theorem and the Class Equation. 
Assume that $G$ only has elements of order $3$. Note that this implies a non-trivial proper subgroup must have order $3$: if there would be such a subgroup $H$ of $G$ order $5$, then an element $x$ inside this subgroup must have order $3$, violating the fact that the order of $x$ must divide $|H|$.
Let $x \in G-Z(G)$, then the cardinality of the conjugacy class $\#Cl_G(x)=|G:C_G(x)|=5$ since by assumption $|C_G(x)|=3$. Hence the Class Equation yields $15=|Z(G)|+5k$ $(*)$, for some non-negative integer $k$. 
If $k=0$, then $G$ would be abelian. Pick $x,y \in G-\{1\}$, $x \neq y$, then ($x$ and $y$ commute) the subgroup $\{1,x,x^2, y, y^2, xy, xy^2, x^2y, x^2y^2\}$ would have order $9$, not dividing $15$. 
Hence $k \gt 0$ and $G$ is not abelian and so $|Z(G)|=1 $ or $3$, but that contradicts equation $(*)$, since $12$ nor $14$ are divisible by $5$.
A: Let me present a correct proof that you would never love, which is adapted from A Proof of Cauchy's theorem.
Assume $G$ is a group of order $15$ with unit $e$. We consider a set $X = \{(x_1,x_2,x_3,x_4,x_5)|\ x_i \in G,\ x_0x_1x_2x_3x_4x_5=e\}$.
Note that the number of elements in $X$, which we denote as $\#X$, is $15^4$, since the first four components determine an element in $X$. Now consider $C_5 = <a>$, the cyclic group of order $5$ with a fixed generator $a$, acts on $X$ as cycles by the following rule: $a$ carries $(x_1,x_2,x_3,x_4,x_5)$ to $(x_2,x_3,x_4,x_5,x_1)$.
Observation 1. According to Orbit-stabilizer theorem, the orbits under the action of $C_5$ should be of length either $1$ or $5$.
Observation 2. Orbits of length $1$ is of the form $\{(x,x,x,x,x)\}$, where $x^5 =e$. In other words, they correspond 1-1 exactly to elements of order $5$ in $G$.
Now by the obvious relation
$$|X| = \#\{\mathrm{orbits\ of\ length}\ 1 \} + \#\{\mathrm{orbits\ of\ length}\ 5 \} * 5,$$ we conclude that $5$ divides $\#\{\mathrm{orbits\ of\ length}\ 1 \}$. Finally, since the unit $e$ is an element of order $5$, i.e. $\{(e,e,e,e,e)\}$ is such an orbit of length $1$, there must be some other length-$1$ orbit, i.e. nontrivial element of order $5$.
