I am reading Dummit and Foote's proof for Cauchy Theorem for abelian groups. On the highlighted line, does anyone know why the authors mention Proposition 2.5(2)?
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.)
The referenced proposition is as follows:
Proposition 2.5(2) Let $G$ be a group, let $x\in G$ and let $a\in \mathbb{Z}-\{0\}.$ If $|x|=n<\infty$, then $|x^a| = n/(n,a)$.
I don't understand how they use this proposition to conclude $p$ divides $|y|$. I think they meant Proposition 2.3, which says
Proposition 2.3 Let $G$ be an arbitrary group, $x\in G$ and let $m,n\in\mathbb{Z}.$ If $x^n=1$ and $x^m=1$, then $x^d=1$, where $d=(m,n).$ In particular, if $x^m=1$ for some $m\in\mathbb{Z}$, then $|x|$ divides $m$.
Can someone confirm?
Proposition 2.5(2) applied to $y$ and $a=p$ gives $|y^p| = \frac{|y|}{(|y|, p)}$. But then $|y| > \frac{|y|}{(|y|,p)}$, since $|y^p| < |y|$ so $(|y|,p) > 1$, however $p$ is prime so $p$ divides $|y|$.
Proposition 2.3 isn't immediately enough since we don't have $y^p = 1$, only that $\bar{y}^p = \bar{1}$ from which we can only conclude that the order of $\bar{y}$ divides $p$, but this was true anyway since the order was $p$.