Fully invariant subgroups The following proposition is a well-known theorem but I don't know where I can find its proof.
Let $G$ be a finite group. Then every subgroup of $G$ is a fully invariant subgroup if and only if $G$ is cyclic group.
Thank you in advance
 A: (1) Sylow subgroups should be fully invariant, hence normal.
(2) Then $G$ is direct product of Sylow subgroups (i.e. it is nilpotent).
(3) If $H$ is a finite $p$-group such that every subgroup is fully invariant, in particular, every subgroup of $H$ is normal in $H$. Try to prove (by induction) that $H$ is either cyclic (if odd order) or possible quaternion (in even order).
[Hint: in induction, note that if a group satisfies a property - being every subgroup is normal, then this property is true also for subgroup as well as quotient, so induction can be applied].
(4) In quaternion group, the subgroup $\langle i\rangle$ and $\langle j\rangle$ can be interchanged by an automorphism: $i\mapsto j, j\mapsto i$. 
(5) So all Sylow subgroups should be normal and cyclic, $G$ should be cyclic.
A: The easy part: If $G$ is cyclic then $\forall T \in End(G),  \exists m \in \mathbb{Z}: \forall g \in G, T(g) = m \cdot g$; so $H \leq G \implies \forall h \in H: T(h) = m \cdot h \in H \implies T(H) \leq H$; i.e. then every subgroup is fully invariant.
Conversely, suppose $G$ is finite, and $\forall H \leq G, \forall T \in End(G): T(H) \leq H$.
Since $Aut(G) \subset End(G)$, this tells us that $\forall H \leq G, g\in G: g^{-1}Hg = H$; i.e., all subgroups of $G$ are normal and so $G$ is a Dedekind group and is either Abelian or has as a direct summand $Q_8$. But $Q_8$ is not fully invariant (as p Groups notes above), so $G$ is finite Abelian.
Suppose $G$ has summands $H \cong Z_{p^a}, K \cong Z_{p^b}$, with $a\leq b$; so that $G \cong H \times K \times B$ for some $B$.
Then $T : (h,k,b) \mapsto (0, p^{b-a}h, b)$ is an endormorpism but $T(H)$ is not a subgroup of $H$.
Therefore there can be only one summand $Z_{p^{n_p}}$ for each $p$ dividing $|G|$, and therefore $G$ is the direct sum of coprime cyclic subgroups, and therefore $G$ is cyclic.
