If a cyclic group G has exactly three subgroups then find order of G.

Question

If a cyclic group $G$ has exactly three subgroups: $G$ itself , {e},and a subgroup of order 7,then find order of G. Please answered me. I tried by Lagrange's theorem.

Answer 1

Since $H$ is a proper subgroup, pick an element $g \in G$ but not in H. But the only subgroups of $G$ are $\{e\}$, $H$ and $G$, and $g \notin \{e\}, H$, so $<g>$ must be $G$ itself. Thus $G$ is cyclic.

Now $7 \mid |G|$ so we can write $|G| = 7k , k \in Z$ . Since $G$ has exactly $3$ subgroups, $|G| $ can have only three positive divisors.

Two of them are $1$ and $7$; if $k = 1$, these are the only two, but then $G$ has only two subgroups. Thus $k >1$ . But if $k > 1$ but not $7$, then $|G|$ has at least four divisors: $1, 7, k$ and $7k$. This would mean that $G$ has at least four subgroups. Thus $k$ must be $7$ and the prime factorization of $|G|$ must be $7^2$, so $|G| = 49$.