# Existence of subgroup of order $p$

I don't understand a part of this lemma.

Lemma. Let $G$ be a finite abelian group of order $m$, let $p$ be a prime number dividing $m$. Then $G$ has a subgroup of order $p$.

Proof. We first prove by induction that if $G$ has exponent $n$ then the order of $G$ divides some power of $n$. Let $b\in G$, $b\neq 1$, and let $H$ be the cyclic subgroup generated by $b$. Then the order of $H$ divides $n$ since $b^n=1$, and $n$ is an exponent for $G/H$. Hence the order of $G/H$ divides a power of $n$ by induction, and consequently so does the order of $G$ because $$(G:1)=(G:H)(H:1).$$ Let $G$ have order divisible by $p$. By what we have just seen, there exists an element $x$ in $G$ whose period is divisible by $p$.

Where is there any hint in the above that there is such an $x$?

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Is this from Lang? I don't think you're alone in finding this proof confusing upon first reading! – Dylan Moreland Aug 13 '11 at 0:28
Yeah, from Lang. Glad I'm not alone. – Matt Aug 13 '11 at 0:54
Definitely not alone! – Alexei Averchenko Aug 14 '11 at 10:02

Suppose that $G$ has no elements of order divisible by $p$. Then every element has order relatively prime to $p$, and thus the lcm of the orders of all elements is itself relatively prime to $p$. Call this $m$. By what you've proven, the order of $G$ must divide a power of $m$. But $p$ divides the order of $G$ (by hypothesis) and $p$ does not divide any power of $m$, a contradiction. Thus, there must be some element of $G$ that has order a multiple of $p$.
Alternatively, let $m$ be the least exponent of $G$; that is, the smallest $k\gt 0$ such that $x^k = 1$ for all $x\in G$. Since the order of $G$ must divide a power of $k$, it follows that $p|k$. Hence, since $k$ is the lcm of the orders of all the elements of $G$, there must be some element of $G$ whose order is a multiple of $p$, as claimed.