# Proof that if $p \mid G$ and $G$ is finite abelian it has an element of order $p$

I want to prove the following lemma.

Let $p$ be a prime number and $G$ a finite abelian group such that $p\mid\#G$, then $G$ has an element of order $p$.

Here's the proof my professor presented.

Let $|G|=pm$ for some $m\in\mathbb{N}$. We will prove the result by induction on $m$.

The base case, $m=1$, is clear (every group of prime order is cyclic).

For the inductive step, choose $x\in G$ such that $\text{ord}(x)=t>1$.

If $p\mid t$ (don't we need to require that $p\neq t$?), then $\text{ord}\big((x^{\frac{t}{p}})^{p}\big)=1$ which implies that $\text{ord}(x^{\frac{t}{p}})=p$, and we are done.

Now suppose that $p\nmid t$. Since $G$ is a abelian then every subgroup is normal, in particular $\langle x\rangle\trianglelefteq G$ and $|G/\langle x\rangle|=\frac{pm}{t}$.

By hypothesis, $\exists y\in G$ such that $\text{ord}(y\langle x\rangle)=p$. (why?) Since for the homomorphism $f:G\rightarrow G/\langle x\rangle$ we have that $\text{ord}\big(f(y)\big)\mid \text{ord}(y)$ it follows that $\text{ord}(y\langle x\rangle)=p\mid \text{ord}\;y$ and we go back to the case where $p$ divides the order or a nontrivial element of $G$.

My questions:

1) the first questions appear in bold in the above text.

2) I don't really get the structure of this induction proof. When the inductive step is presented I don't see in what way we assume that the results holds for $m$ and how we show it is true for $m+1$.

My apologies if these questions are too basic but I am a former physics student and lack some background and experience with proofs.

• Is $p$ a prime number? – paw88789 Oct 17 '17 at 16:20
• Yes, I forgot to include this. Thanks. – aadcg Oct 17 '17 at 16:21
• For the first bold part: Not needed since if $p=t$ all the statements that follow hold by definition of $x$. $ord((x^{t/p})^p)=ord(x^p)=ord(e)=1$. For the second bold part: Observe that $G/(x)$ has order $p(m/t)$ and $m/t < m$. Therefore, $G/(x)$ has an element of order $p$. That element is being called $y(x)$, the coset of some element $y\in G$. – Hellen Oct 17 '17 at 16:29
• In finite abelian groups,Converse of Lagrange's theorem holds good. – Sumit Mittal Oct 17 '17 at 17:26

don't we need to require that $p \neq t$?

Not that I can see from your argument. If $\operatorname{ord}(x) = p$, then there is nothing to prove. That said, the way you currently have structured your argument, you will need to ensure $p \nmid t$ when applying the inductive hypothesis to the group $G / \langle x \rangle$.

One way around this would be to replace your current definition of $x$ with something more specific. For instance, you could say

Choose any $x \in G \setminus \{ 1 \}$. If $p \mid \operatorname{ord}(x)$, say $\operatorname{ord}(x) = pq$ for $q \in \mathbb{N}$, then by definition $y = x^q \in G$ is an element of order $p$. Conversely, suppose $p \nmid \operatorname{ord}(x)$ [$\ldots$]

and take things from there. This seems to be what you mean, but calling out the two possible cases more explicitly goes a long way towards clarifying the rest of the argument (for both the reader and you!).

By hypothesis, $\exists y \in G$ such that $\operatorname{ord}(y\langle x\rangle)=p \operatorname{ord}(y\langle x\rangle)=p$. (why?)

If you assume $p \nmid t$, then this is because $|G/\langle x \rangle|$ is of the form $pm'$ where $m' < m$ (and so the inductive hypothesis applies).

Regarding your last question: It sounds like you're assuming that this is a proof by weak induction, when this is really a proof by strong induction. Suppose $P$ is a property of the natural numbers, and you know that $P(1)$ is true. The difference in proving that $P(N)$ is true for all $N \in \mathbb{N}$ using weak and strong induction is the following:

• Using weak induction, you assume for a fixed $N \in \mathbb{N}$ that $P(N)$ holds and show that this implies $P(N + 1)$.
• Using strong induction, you assume for a fixed $N \in \mathbb{N}$ that $P(n)$ holds for all $n \leq N$ and show that these statements together imply $P(N + 1)$ holds.

See this set of notes for examples of both types of proofs. Each of these is a valid form of induction.

Your proof relies on strong induction, and you can see this from how you apply your inductive hypothesis. First, you choose an arbitrary $x \in G$ with $x \neq 1$ and $p \nmid t := \operatorname{ord}(x)$. You then apply the inductive hypothesis to the group $G/\langle x \rangle$. Supposing that $|G/\langle x \rangle| = pk$, you don't know the value of $k$, since you don't know enough about $x$. All you know is that it is less than $m$ where $|G| = pm$. To cover all bases, you have to assume the claim holds for all finite abelian groups of order $pn$ for $n < m$, not just for those of order $p(m - 1)$.

• Very clear and informative. – aadcg Nov 10 '17 at 0:25

1) If $p=t$, $t/p=1$, so $x$ has order $p$ anyway and is the element we want.

2) Your second piece of bold is where the inductive step is used: the key point is that $G/\langle x \rangle$ is a group of order $pm/t$, and if $p \not\mid t$, then there is an integer $n$, smaller than $m$, so that $pm/t = pn$. The induction hypothesis then says the group $G/\langle x \rangle$ has an element of order $p$; and then the structure of the quotient group means that there is $y \in G$ so that this element can be written as $y\langle x \rangle$.