# 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 $p$.

• I think the best thing to do here would be to prove cauchy – Jorge Fernández Hidalgo Sep 22 '14 at 23:43
• Yes, but I think the proof of this without Cauchy must reveal something interesting about the order of quotient groups compared to the order of their elements, which does hold for non-quotient groups. – MPitts Sep 22 '14 at 23:55
• This implies Cauchy's theorem, just taking $a$ to be the identity of $G$, so there's no real way to get around it. – Kevin Carlson Sep 23 '14 at 0:16

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:

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

2. 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.)

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

4. 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 $g$?

• Sorry if I'm being dense here, but there are two aspects of the proof and explanation that I don't see: (1) where did we prove that the proposition is true for k if it is true for values less than k ? And (2) problem 2 concerns the order of the class G/<a>, whereas problem 3 concerns the order of an element (in this case a coset) in that class. Obviously, these can be two very different orders. – MPitts Sep 23 '14 at 4:17
• @user54301: (1) We are doing precisely that in problems 1-4. We showed (trivially) that it's true for $k=1$, then, during all four problems, we assume that it's true for all groups with order less than $k$. The goal of the problems is to prove that this implies it's true for $k$. Since $G/\langle a\rangle$ has order less than $k$ and divisible by $p$ (not necessarily equal to $p$), the proposition is true for that group: namely, that group contains an element of order $p$. – Bungo Sep 23 '14 at 4:35
• Ok, I now see that problems 1-4 as a whole are the proof by induction. But can you help me see why problem (3) is true? – MPitts Sep 23 '14 at 21:41
• In the explanation of (3), you mention induction on |G/<a>|, but doesnt the proof instead concern an induction on k = |G| ? – MPitts Sep 23 '14 at 21:55
• @user54301: The induction hypothesis is that the proposition is true for any finite group which has order less than $k$ and divisible by $p$. – Bungo Sep 23 '14 at 21:57

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 $p$.