$G$ characterstically simple $\rightarrow$ $pG=0$ for all primes $p$ or $pG=G$ for all primes p $G$ is abelian
$G$ characterstically simple $\rightarrow$ $pG=0$ for all primes $p$ or $pG=G$ for all primes p
I understand that for a certain prime $p$ why if $G$ is characteristically simple then $pG=0$ or $pG=G$... but how come this holds for ALL primes?
Why can't there be primes $p_1,p_2$ s.t. $p_1G=0$ and $p_2G=G$?
Thanks!
 A: As written, your statement is wrong unless you are only talking about abelian groups $G,$ but that is (almost) implied because $pG$ is rarely written for non-abelian groups.
Also, the statement should be:

For all primes $p,$ either $pG=\{0\}$ or $pG=G.$

That is not the same as what you have written:

Either for all primes $p,$ $pG=\{0\}$ or for all primes $p$, $pG=G.$

Your statement does not allow one prime $p$ with $pG=\{0\}$ and the rest of the primes $p$ with $pG=G.$ As such, your statement is actually false.
Indeed, there can never be more than one prime $p$ such that $pG=0$ unless $G=0,$ as I show in the aside below.

Answer
I will show:

If $G$ is an abelian characteristically simple group, then for any prime $p$ either $pG=\{0\}$ or $pG=G.$

Given any abelian $G,$ and any $n$ natural, first show that $nG$ is a characteristic subgroup - that is, given any automorphism $\phi:G\to G$, $\phi(nG)=nG.$ (Proof left to you.)
If $G$ is characteristically simple, then any characteristic subgroup of $G$ must be either $\{0\}$ or $G.$ 
So we have the stronger result - if $G$ is abelian and characteristically simple, then for any natural number $n,$ either $nG=\{0\}$ or $nG=G.$
In particular, this is true for any prime $n=p.$

Aside
Claim: If $G$ is a non-trivial abelian group, there can be at most one prime $p$ such that $pG=0.$
Proof: Given two primes $p_1,p_2$ with $p_1G=\{0\}=p_2G$ then either $G=\{0\}$ or $p_1=p_2.$
This is because if $p_1\neq p_2$ then there are integers $x,y$ such that $p_1x+p_2y=1$ and then for any $g\in G,$ we get: $$g=1\cdot g=(p_1x)\cdot g)+(p_2y)\cdot g = p_1\cdot(x\cdot g)+p_2\cdot(y\cdot g)=0+0=0.$$
So if $G\neq\{0\}$ is abelian, there can be at most one prime $p$ such that $pG=\{0\}.$ 
