0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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! $\endgroup$ – Dietrich Burde Oct 28 '16 at 9:11
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Will R Oct 28 '16 at 8:57
  • 1
    $\begingroup$ 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 $6$. $\endgroup$ – Leppala Oct 28 '16 at 9:06
  • $\begingroup$ I guess I got the names of the theorems mixed up, let me edit. $\endgroup$ – Servaes Oct 28 '16 at 9:16
1
$\begingroup$

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 $\text{order}(x)$.

$\endgroup$
  • $\begingroup$ Thank you for your answer. But how do I know that $ord(x)|ord(H)$. In our lessons, the Lagrange-Theorem only says: $ord(G)=ord(H)*(G:H)$. $\endgroup$ – milui Oct 28 '16 at 8:57
  • $\begingroup$ thank you very much, now I understand $\endgroup$ – milui Oct 28 '16 at 9:00
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.