1

Let $G$ be a group with $|G|>1.$ Suppose $G$ and $\{e_G\}$are only subgroups of $G.$ Then, there exists $p \in \mathbb{P}$ such that $G \cong \mathbb{Z}/(p).$

May I know if my proof is correct? Thank you for your attention.

Proof: Let $|G|= p \in \mathbb{N}.$ Suppose $p \not \in \mathbb{P},$ then there exists $d \in \mathbb{P}$ such that $d|p.$ By Cauchy's theorem, $\exists q \in G: o(q) = d,$ whence $1<| \left \langle {q}\right \rangle| =d < p.$ (Contradiction).

Hence, $G$ is cyclic and $G = \left \langle {g}\right \rangle,$ for some $g \in G.$ Given $g^i \in G,$ let $g^i \mapsto \ [1^i]_p \in (\mathbb{Z}/(p),+), $ The mapping is an isomorphism.

Alexy Vincenzo
  • 5,400

1 Answers1

2

It looks fine (although I think you meant to say $o(q) = d$, not $o(q) = p$), but you don't need to use Cauchy's theorem if you don't want to: choose $g\in G$, $g\neq e$. Then $\langle g\rangle$ is a nontrivial subgroup of $G$, but the only subgroups of $G$ are $G$ and $\{e\}$. Therefore, $\langle g\rangle = G$. Now, if $\left|G\right|$ were not prime, say $\left|G\right| = n\cdot m$, then $\langle g^{n}\rangle$ would be a nontrivial subgroup of $G$, which is a contradiction. Thus, $G$ is cyclic of order $p$ (for some prime $p$), so $G\cong\Bbb Z/(p)$ (via the isomorphism you described).

Stahl
  • 23,212