A group G with order $15$ is simple?

Question

A group $G$ with order $15$ is simple? There are a theorem for realize it? Thanks for all you help!

Answer 1

There are many arguments to prove that group of order $15$ is not-simple (and in fact, is cyclic).

I came, few years before, across the following interesting argument, through an exercise from Gallian's "Contemporary Abstract Algebra". (At that time, the exercise taught me- Very elementary facts can be nicely used to crack some simple statements in Algebra)

Let $|G|=15$. By Cauchy's theorem, there is an element (subgroup) of order $5$. Let $H$ denote a subgroup of order $5$.

If $K\neq H$ is another subgroup of order $5$, then $K\cap H=1$ (why?), hence $$|HK|=\frac{|H|\cdot|K|}{|H\cap K|}=\frac{5\cdot5}{1}>|G|,$$ a contradiction. Hence subgroup of order $5$ is unique, hence normal, hence $G$ is not simple.

Answer 2

The cyclic group $\{e,a,a^2,a^3,\ldots,a^{14}\}$ of order $15$ is not simple. The set $\{e,a^5,a^{10}\}$ is a normal subgroup.