I am reading the Dummit and Foote's proof of Cauchy's theorem for abelian groups. On the highlighted line, how do you conclude from $y\notin N$ and $y^p\in N$ that $\langle y^p \rangle \neq \langle y \rangle$ and that $|y^p|<|y|$?
Proposition 21. If $G$ is a finite abelian group and $p$ is a prime dividing $|G|$, then $G$ contains an element of order $p$.
Proof: The proof proceeds by induction on $|G|$, namely, we assume the result is valid for every group whose order is strictly smaller than the order of $G$ and then prove the result valid for $G$ (this is sometimes referred to as complete induction). Since $|G| > 1$, there is an element $x \in G$ with $x \neq 1$. If $|G| = p$ then $x$ has order $p$ by Lagrange’s Theorem and we are done. We may therefore assume $|G| > p$.
Suppose $p$ divides $|x|$ and write $|x| = pn$. By Proposition 2.5(3), $|x^n| = p$, and again we have an element of order $p$. We may therefore assume $p$ does not divides $|x|$.
Let $N = \langle x \rangle$. Since $G$ is abelian, $N \trianglelefteq G$. By Lagrange’s Theorem, $|G/N| = |G|/|N|$ and since $N \neq 1$, $|G/N| < |G|$. Since $p$ does not divide $|N|$, we must have $p \mid |G/N|$. We can now apply the induction assumption to the smaller group $G/N$ to conclude it contains an element $\bar{y} = yN$, of order $p$. Since $y \notin N$ ($\bar{y} \neq \bar{1}$) but $y^p \in N$ ($\bar{y}^p = \bar{1}$) we must have $\langle y^p \rangle \neq \langle y \rangle$, that is, $|y^p| < |y|$. Proposition 2.5(2) implies $p \mid |y|$. We are now in the situation described in the preceding paragraph, so that argument again produces an element of order $p$. The induction is complete.
(Original image here.)
Saying that $y^p\in N$ is the same as saying $\langle y^p\rangle\le N$. Then if $\langle y^p\rangle=\langle y\rangle$ you have $\langle y\rangle\le N$, so $y\in N$.
Now the second point follows from the first, since $\langle y^p\rangle\le\langle y\rangle$ (since $y^p\in\langle y\rangle$) but $\langle y^p\rangle\neq\langle y\rangle$.