# Group of order $pq$. Show that there exist an element of order $p$ in G.

I don't know how to start. In lessons we didn't heard about the Slylov-Theorem yet.

Let $q$ and $p$ be a prime number and $G$ an abelian group of order $pq$ ($|G|=pq$). Show that there exists an element of order $p$ in $G$.

Now I would start with the Lagrange-Theorem. So: $\mathrm{ord}(G)=\mathrm{ord}(H)*(G:H)$. I know that there exists subgroups of order $p$ and $q$ but what does this tell me about the order of an element in $G$.

Thank you for taking your time.

• Cauchy's Theorem is certainly worth to be considered here. It would give us an element of order $p$ and also an element of order $q$. It can be proved without Sylow. By the way, how would you know that there exists a subgroup of order $p$ in $G$? This is the whole point! – Dietrich Burde Oct 28 '16 at 9:11

We can do this directly. By Lagrange's theorem, the order of every element divides the order of $G$, which is $pq$. The only divisors of this are $pq, p, q$, and 1.

Let us work through each of these separately.

Suppose that there exists an element of order $pq$, call it $x$. Then $x^q$ has order $p$, and you are done.

Suppose next that there is an element $y$ of order $p$. Done!

Tricky case: Suppose now that we have an element $z$ of order $q$. Then we can examine the group $G/\langle z \rangle$, which has order $pq/q = p$. Choose a generator of this, and lift it to $G$. Then this has either order $pq$ or order $p$ and we are in one of the first two cases.

Lastly, suppose that all elements have order one. This can't occur, since the order of $G$ is $pq$, and so we are done.

You already know there exists a subgroup $H\subset G$ of order $p$. What are the orders of the elements of $H$? Remember that every element of $H$ generates a subgroup of $H$, so you can apply Lagrange's theorem here.

• I don't see how the existence of a subgroup of order $p$ can be deduced from Lagrange's theorem. I would refer typically refer to Cauchy's theorem for this. Am I just missing something? – Will R Oct 28 '16 at 8:57
• As @WillR said, Lagrange only says that IF $H \leq G$, THEN $|H|$ divides $|G|$. The converse in not true in general. For example $A_4$ (of order $12$) doesn't have a subgroup of order $6$. – Leppala Oct 28 '16 at 9:06
• I guess I got the names of the theorems mixed up, let me edit. – Servaes Oct 28 '16 at 9:16

Assuming that like you said you already know that there exists a sub-group $H$ of order $p$.

Then take $x\in H$, by Lagrange theorem, $\mathrm{order}(x)\mid \mathrm{order}(H)$. Since $p$ is prime, is only divisors are $1$ and $p$.

You just need to take $x\ne 1_H$ and you are done.

Edit

Why $\mathrm{order}(x)\mid \mathrm{order}(H)$ ?

Because take the sub-group of $H$ generated by $x$ : $\langle x\rangle$. It is a subgroup of $H$ of cardinal $\text{order}(x)$.

• Thank you for your answer. But how do I know that $ord(x)|ord(H)$. In our lessons, the Lagrange-Theorem only says: $ord(G)=ord(H)*(G:H)$. – milui Oct 28 '16 at 8:57
• thank you very much, now I understand – milui Oct 28 '16 at 9:00

You may also like to use Cauchy's theorem.https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)

There is a subgroup of order $p$ p is prime. A Group of prime order is cyclic. A cyclic group of order $p$ has an element of order $p$.