I've been trying to show that any group of order $15$ is cyclic and the only missing part in my proof is to show that the subgroup with $3$ elements (which we know exists by Cauchy's theorem) is actually normal.
I've shown the normality of the subgroup of order $5$ by its uniqueness but such a trick does not work with $3$ because $3$ is too small. I know by the 3rd Sylow theorem that $N\equiv 1 \mod 3$ where $N$ is the number of subgroups of order $3$ and since $N\mid 15$ it must be $1$. So this would imply that the subgroup is unique hence normal.
But is there a way to show this result without resorting to Sylow?
Best Answer
If we know that in a group $G$ of order $15$ a subgroup of order $5$ is normal, then it follows easily that $G$ is a cyclic group.
Indeed let $a$ and $b$ be elements of the group $G$ of order $3$ and $5$ respectively. Since the subgroup $\langle b\rangle$ is normal in $G$, we have $$ a^{-1}ba=b^k, $$ where $k$ is one of the numbers $\{1,2,3,4\}$. Next $$ b=a^{-3}ba^3=a^{-2}b^ka^2=(a^{-2}ba^2)^k=\ldots=b^{k^3}. $$ Hence $k^3\equiv1\pmod5$. It follows that $k=1$. The rest is clear.