Every group of order $70$ has a normal subgroup isomorphic to $\mathbb{Z}_{35}$?

Question

Show that every group of order $70$ has a normal subgroup isomorphic to $\mathbb{Z}_{35}$

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I'm not sure if I can use Sylow's theorems here, since this is asking about a group whose order is not a power of a prime.

Answer 1

You can find a proof here that there exists a normal subgroup of order $35$. It is isomorphic to $\mathbb Z_{35}$ as every group of order $35$ is cyclic (since its order if of the form $pq$ with $p<q$ and $p \not \mid q-1$).

You can also find another proof of the fact that every group of order 35 is cyclic, here.

Answer 2

Show that every group of order 70 has a normal subgroup isomorphic to $\mathbb{Z}_{35}:$

You can still use Sylow's theorems here. Let $n_7$ be the number of 7-subgroups of Sylow (as $70=2\times 3\times 7$). Then $n_7|10$, and $n_7\equiv 1\;(mod\;7)$, the only choice you have is $n_7=1$. Now, for Sylow's second theorem we know that the 7-subgroup $P_7\triangleleft G$. Moreover we know that if $N\triangleleft G$ and $K\leq G$ then $NH\leq G$. We apply this, so $P_7P_5\leq G$ (we know $P_5$ exists for Sylow's first theorem). Notice that the $P_7P_5$ index is 2 (70/35), what implies $P_7P_5\triangleleft G$.

To conclude you can use that every group of order $pq$ with $p>q$ and $q\nmid p-1$ is cyclic, and so is $P_7P_5$, which order is $35=5\times 3$ where $3\nmid 5-1$.

$$\exists H\triangleleft G:H\cong\mathbb{Z}_{35}$$