Showing $G$ is the item of teams of prime order
Let $G$ be a (not always limited) team with the building that for each and every subgroup $H$ of $G$, there exists a ' retraction ' of $G$ to $H$ (that is, a team homomorphism from $G$ to $H$ which is identification on $H$). After that, we assert:

$G$ is abelian.

Each component of $G$ has limited order.

Each component of $G$ has square  free order.
Allow $g$ be a nontrivial component of $G$ and also take into consideration a retraction $T : G \to \langle{g\rangle}$ which is identification on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic therefore, it is abelian.
Apart from this i do not recognize just how to confirm the various other cases of the trouble. In addition, a comparable trouble was asked in Berkeley Ph.D test, in the year 2006, which in fact asks us to confirm that:
If $G$ is limited and also there is a retraction for each and every subgroups $H$ of $G$, after that $G$ is the items of teams of prime order.
Let $H$ be a subgroup of $K$ which is a subgroup of $G$. If there is a retraction of $G$ onto $H$, it limits to a retraction of $K$ onto $H$. So if a team $G$ has this building (allow is claim it is "retractible") after that each subgroup of $G$ is retractible. Which cyclic teams are retractible?
Let $g$ be a nontrivial component of $G$ and also take into consideration a retraction $T : G \to \langle{g\rangle}$ which is identification on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic therefore, it is abelian.
Hence $[G,G]$ is had in $Ker(T)$. Given that $g \notin Ker(T)$, $g \notin [G,G]$. As $g$ is an approximate nontrivial component of $G$, this suggests that $[G,G] = {e}$ ; that is, $G$ is abelian.
Consider any kind of component $g \in G$ and also take into consideration a retraction $T:G \to \langle{g^2 \rangle}$. $T(g)$ remains in $\langle{g^2 \rangle}$ suggests $T(g) = g^{2r}$ for some $r$. Additionally, $T(g^2)=g^2$ suggests then that $g^{4r}=g^2$ ; that is, $g^{4r2} = e$. As $4r2$ is not absolutely no, we get that $g$ has limited order.
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