Let $G$ be an abelian finite non trivial group and $p$ a prime number which devises o(G). We denote $H$ the set of non trivial subgroups of $G$. Could you please help me to show that if there exists $M\in H$ such that $p$ devises $o(M)$  then there exists $g\in G$ of order $p$. Thanks in advance
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 $|y^p|&lt;|y|$?     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 $|G| &gt; 1$, 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 $|G| &gt; p$.      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 $N \neq 1$, $|G/N| &lt; |G|$.   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, $|y^p| &lt; |y|$.   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 .)
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