Question

Show that every group of order 5*7*47 is abelian and cyclic.

Answer 1

We will now show that every group of order 5*7*47=1645 is abelian, and cyclic by Theorem 1:

The group $\Z_m \times \Z_n$ is isomorphic to $\Z_{mn}$ if and only if $gcd(m,n)=1$ .

By the Third Sylow Theorem, $G$ has only one subgroup $H_1$ of order 47. So $G/H_1$ has order 35 and must be abelian by Theorem 2:

If $p$ and $q$ are distinct primes with $p<q$ , then every group $G$ of order $pq$ has a single subgroup of order $q$ and this subgroup is normal in $G$ . Hence, $G$ cannot be simple. Furhermore, if $q\not \equiv 1$ (mod $p$), then $G$ is cyclic.

Hence, the commutator subgroup of $G$ is contained in $H$ which tells us that $|G'|$ is either 1 or 47. If $|G'|=1$ , we are done. Suppose that $|G'|=47$. The Third Sylow Theorem tells us that $G$ has only one subgroup of order 5 and one subgroup of order 7. So there exist normal subgroups $H_2$ and $H_3$ in $G$ , where $|H_2|=5$ and $|H_3|=7$ . In either case the quotient group is abelian; hence, $G'$ must be a subgroup of $H_i, i=1,2$ . Therefore, the order of $G'$ is 1, 5, or 7. However, we already have determined that $|G'|=1$ or $47$ . So the commutator subgroup of $G$ is trivial, and consequently $G$ is abelian.