Order of elements of quotient group How do we prove the following without using Cauchy's theorem? (From Pinter p156 problem H3.)
Let $G$ be a finite abelian group, and the order $|G|$ be a multiple of a prime number, $p$.
Let $a$ be an element of $G$ and $\text{ord}(a)$ not a multiple of $p$. It can be shown that the order of the quotient group $|G/\langle a \rangle|$ is a multiple of $p$.
Prove $G/\langle a\rangle$ must have an element of order $p$.
 A: Looking at page 156, I see that this question is in the context of "induction on $|G|$: an example." The series of four exercises is walking you through a proof of Cauchy's theorem for the special case of finite abelian groups, using induction.
For the benefit of others reading this question, I'll summarize the context here.
Claim: if $G$ is a finite abelian group, and $p$ is any prime factor of $|G|$, then $G$ has an element of order $p$. The base case, $|G| = 1$, is trivially true since there are no prime factors of $|G|$. Assume that $|G| = k$ and the result is proven for orders smaller than $k$. Take any nonidentity element $a \in G$. If $\text{ord}(a) = p$, then we're done. Otherwise:


*

*If $\text{ord}(a) = tp$ for some positive integer $t$, what element of $G$ has order $p$?

*Suppose $\text{ord}(a)$ is not a multiple of $p$. Then $G/\langle a \rangle$ is a group having fewer than $k$ elements. (Explain why.) The order of $G/\langle a \rangle$ is a multiple of $p$. (Explain why.)

*Why must $G / \langle a \rangle$ have an element of order $p$?

*Conclude that $G$ has an element of order $p$.
I assume you have answered 2 and now want to know why 3 is true. But this is simply because $|G / \langle a \rangle| < k$ and $p$ divides $|G / \langle a \rangle|$ (by problem 2), and we are assuming that we have proved the result for finite abelian groups of order is less than $k$ and divisible by $p$. (Induction hypothesis.)
By the way, problem 2 doesn't mention it, but for completeness you need to show that $G/\langle a \rangle$ is abelian. Fortunately, this is easy.
Problem 4 is actually the key part of the proof. Do you see why it's true? Hint: by 3, there is an element of $G/\langle a \rangle$ with order $p$. The elements of this group are cosets of $\langle a \rangle$, so let's call the element $g\langle a\rangle$. If $g\langle a\rangle$ has order $p$, then what does that tell you about the order of $g$?
A: As explained by Kevin Carlson, this implies Cauchy's Theorem anyway. So let's prove Cauchy's Theorem for finite abelian groups.
Write $G = \bigoplus_q G_q$ with $G_q = \{a \in G : \exists n \geq 0 ( q^n a = 0)\}$ for primes $q$. Then $|G_q|$ is a power of $q$, because any element $a \in G_q$ has order a power of $q$ and we can do induction by considering $G_q / \langle a \rangle$. Now if a prime $p$ divides $|G|$, it divides some $|G_q|$. By the result before, we must have $q=p$. This means $G_p \neq 1$, i.e. there is some element $a$ of order $p^n$, $n \geq 1$. Then $p^{n-1} a$ has order $p$.
