Group of order 105 as a product

Question

I can't solve this problem, please help me!

Every group of order 105 is isomorphic to $\mathbb{Z}_5\times H$ where $H$ is a group of order 21.

Answer 1

Note that $105 = 3\cdot 5\cdot 7$. Now think about Sylow's theorems. If $n_7=1$, we're done since we have a subgroup of order $3\times 7$. If $n_7=15$, we have $15\times 6=90$ elements of order $7$. We're then left with only $15$ elements. But $n_5=1,21$. So $n_5=1$. But then we have only one $3$-Sylow, and hence we again have a subgroup of order $3\times 7$. It remains that you show that if $H$ is a subgroup of $G$ of order $21$, $H\lhd G$, and that $n_5=1$.

(Recall that if $H\lhd G$ and $K\leqslant G$, $HK\leqslant G$, and $|HK|=|H||K|/|H\cap K|$.)