Group with order $30$ is not a simple group.

Question

Could someone prove the sentence given above in the title?

I know that Sylow theorems should be used here.

Let $N_{p}$ stands for number of $p$-Sylow subgroups in group $G$.

I tried to use following sentences:

• $\lvert G \rvert = 30 =2 \cdot 3 \cdot 5$
• $N_{p} \equiv 1 \pmod p$
• $N_{p} \mid 30$

But i can't figure out.

Regards.

Answer 1

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$If you know the result

if the finite group $G$ has order $\Size{G} = 2 d$, with $d$ odd, then $G$ has a subgroup of order $d$,

then you are immediately in business.

Otherwise, check the possibilities for the number of Sylow $5$-subgroups, and consequently for the number of elements of order $5$. If $N_{5} > 1$, you will find that there won't be too many elements left.