# Showing $G$ is the product of groups of prime order

Let $G$ be a (not necessarily finite) group with the property that for each subgroup $H$ of $G$, there exists a `retraction' of $G$ to $H$ (that is, a group homomorphism from $G$ to $H$ which is identity on $H$). Then, we claim :

• $G$ is abelian.

• Each element of $G$ has finite order.

• Each element of $G$ has square-free order.

Let $g$ be a nontrivial element of $G$ and consider a retraction $T : G \to \langle{g\rangle}$ which is identity on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic and so, it is abelian.

Other than this i don't know how to prove the other claims of the problem. Moreover, a similar problem was asked in Berkeley Ph.D exam, in the year 2006, which actually asks us to prove that:

If $G$ is finite and there is a retraction for each subgroups $H$ of $G$, then $G$ is the products of groups of prime order.

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There is no "Berkeley Ph.D. exam". You are talking either about the Berkeley Preliminary examinations (part of Ph.D. program, they appear in the book "Berkeley Problems in Mathematics") or a problem from a Qualifying Exam in Berkeley. – Arturo Magidin Sep 29 '10 at 16:35
@Arturo: Thats what i meant! – anonymous Sep 29 '10 at 20:04
@Chandru: I mentioned two (mutually exclusive) options. Apparently, you meant one of them. Sadly, you neither specified which one you meant, nor fixed the assertion in the question. – Arturo Magidin Sep 29 '10 at 22:33

Let $H$ be a subgroup of $K$ which is a subgroup of $G$. If there's a retraction of $G$ onto $H$, it restricts to a retraction of $K$ onto $H$. So if a group $G$ has this property (let's say it's "retractible") then each subgroup of $G$ is retractible. Which cyclic groups are retractible?

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I have worked out some more details. why do we need the fact "Which cyclic groups are retractible" :x) – anonymous Sep 29 '10 at 14:51
Now i am stuck up on the last part :)! – anonymous Sep 29 '10 at 14:52
@Chandru: you may perhaps not "need" to know which cyclic groups are retractable, but if you did know which cyclic groups have the desired property, then you would have easily obtained the second and third part of the problems as a consequence. For example, the infinite cyclic group is not retractable, because all nontrivial images are finite, but all nontrivial subgroups are infinite (which immediately gives you the second property). The third part of the problem will follow if you figure out which finite cyclic groups have the desired property. – Arturo Magidin Sep 29 '10 at 23:11

Let $g$ be a nontrivial element of $G$ and consider a retraction $T : G \to \langle{g\rangle}$ which is identity on $\langle{g\rangle}$. As $G/Ker(T)$ is isomorphic to $\text{Img}\langle{g\rangle}$, it is cyclic and so, it is abelian.

Thus $[G,G]$ is contained in $Ker(T)$. Since $g \notin Ker(T)$, $g \notin [G,G]$. As $g$ is an arbitrary nontrivial element of $G$, this means that $[G,G] = {e}$; that is, $G$ is abelian.

Look at any element $g \in G$ and consider a retraction $T:G \to \langle{g^2 \rangle}$. $T(g)$ is in $\langle{g^2 \rangle}$ means $T(g) = g^{2r}$ for some $r$. Also, $T(g^2)=g^2$ means then that $g^{4r}=g^2$; that is, $g^{4r-2} = e$. As $4r-2$ is not zero, we get that $g$ has finite order.

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