A question in Abelian Group

Question

Suppose $G$ is a finite Abelian group, and $U$ is a cyclic subgroup of $G$ with the maximal order. Choose $y$ in $G-U$ such that $|y|$, the order of y, is the minimal. Suppose $p$ is a prime number and divides $|y|$, then prove that $<y^p>\le U$. This is one step in proving $U$ has a compliment in $G$, and I just don't get this step. Why $y$ is not an element in $U$ but a subgroup $<y^p>$ is in $U$. I mean if $y$ is not in $U$, then $<y>$ wouldn't be $U$ is this correct? If so, how come a subgroup of $<y>$ is in $U$?

Answer 1

That's actually very easy.

$<y>$ isn't in $U$, but it doesn't mean that they are disjoint.
For example, in $\mathbb{Z}/\mathbb{4Z}$, take $U=\{0,2\}$, and $y=1$.
you have that $y\not\in U$, but $<y>=\mathbb{Z}/\mathbb{4Z}$, so it's not disjoint from $U$.
Moreover, $<2y>=U$

Answer 2

The comment suggests that you may already have an answer, but here it is for the record.

The element $y^p$ has smaller order than $y$ (namely the order of $y$ divided by$~p$). Therefore by the choice of $y$, one cannot have $y^p\in G-U$, so $y^p\in U$. But since $U$ is a subgroup this imples $\langle y^p\rangle\subseteq U$.