A Question about p-primary Group For a p-primary group $G$, if  $\langle y\rangle$ is a cyclic group of order $p^i$ for any natural integer $i$, then $ng \in \langle y\rangle$ implies $g \in \langle y\rangle$ for any $g \in G$?
Also, in the terms of group theory, what does the symbol $G^p$ means for a group $G$?
I wanted to ask this because I was referring to this post:Example of Intersection of Pure Subgroup which is not Pure
Note that $G$ is abelian. Thanks in advance.
 A: The answer to your first question is "no". Take $G=C_p\times C_{p^2}$, a product of a cyclic group of order $p$ and a cyclic group of order $p^2$, and let $y=(1,1)$, which has order $p^2$. Then let $g=(0,1)$. Then $pg = (0,p) = p(1,1) = py$, so $pg\in \langle y\rangle$, but $g\notin \langle y\rangle$.
Added for completeness: As mentioned in the comment below, if you also have that $\gcd(n,p)=1$, however, then the conclusion does follow: since $G$ is a $p$-group, there exists $k$ such that $p^kg = 0$, and from $\gcd(n,p)=1$ we know there exist integers $r,s$ such that $1 = rn + sp^k$. Then, since $ng\in\langle y\rangle$, we have:
$$g = 1g = (rn+sp^k)g = (rn)g + (sp^k)g = r(ng) + s(p^kg) = r(ng)+s0 = r(ng)\in\langle y\rangle.$$
Note that the above works in any group: if $H$ is a subgroup of $G$, $g\in G$ is of order $k$, and $g^n\in H$ with $\gcd(n,k)=1$, then $g\in H$, essentially  because $\langle g\rangle = \langle g^n\rangle$ if $\gcd(n,k)=1$, so $$g\in\langle g\rangle = \langle g^n\rangle \subseteq H.$$
For your second question: for a group $G$, $G^p$ is the subgroup of $G$ generated by all elements of the form $x^p$ with $x\in G$; that is, the subgroup generated by the $p$th powers of all elements of $G$.
If $G$ is abelian, then this is simply the subgroup that consists of all $p$-powers, since $a^pb^p = (ab)^p$; for nonabelian groups, it may in general be larger.
Your first sentence is not very well-written, by the way: "any" is incorrect there, you mean "some" (otherwise, you are saying that for all natural numbers $i$, the subgroup $\langle y\rangle$ has order $p^i$; this is impossible, since the subgroup can only have one order).
