1. **Problem Statement:** Prove that all subgroups of a cyclic group are fully invariant subgroups of that cyclic group. 2. **Definitions:** - A **cyclic group** $G$ is a group generated by a single element $g$, i.e., $G = \langle g \rangle$. - A **subgroup** $H$ of $G$ is a subset of $G$ that is itself a group under the operation of $G$. - A subgroup $H$ is **fully invariant** if for every endomorphism $\varphi: G \to G$, we have $\varphi(H) \subseteq H$. 3. **Key fact about subgroups of cyclic groups:** Every subgroup $H$ of a cyclic group $G = \langle g \rangle$ is also cyclic and can be written as $H = \langle g^d \rangle$ for some divisor $d$ of the order of $G$ (or $d \geq 0$ if $G$ is infinite). 4. **Proof:** - Let $G = \langle g \rangle$ be a cyclic group and $H = \langle g^d \rangle$ a subgroup of $G$. - Consider any endomorphism $\varphi: G \to G$. Since $G$ is cyclic, $\varphi$ is completely determined by $\varphi(g) = g^k$ for some integer $k$. - Then for any element $h \in H$, $h = (g^d)^m = g^{dm}$ for some integer $m$. - Applying $\varphi$ to $h$ gives: $$\varphi(h) = \varphi(g^{dm}) = (\varphi(g))^{dm} = (g^k)^{dm} = g^{kdm}.$$ - Since $H = \langle g^d \rangle$, to show $\varphi(h) \in H$, we need $g^{kdm} \in \langle g^d \rangle$, which means $d$ divides $kdm$. - Because $d$ divides $dm$, it suffices to check that $d$ divides $kdm$, which is true for all integers $k,m$. - Therefore, $\varphi(h) \in H$ for all $h \in H$ and all endomorphisms $\varphi$. 5. **Conclusion:** All subgroups of a cyclic group are fully invariant because every endomorphism of the cyclic group maps each subgroup into itself. **Final answer:** All subgroups of a cyclic group are fully invariant subgroups of that cyclic group.