1
$\begingroup$

This is question 13.2 in Armstrongs "Groups and Symmetry". I've googled quite a bit but I can't find an overview of the solutions of the the book. Variations of this particular question come back in multiple instances and I can't really figure it out.

"If $p_1,p_2,...p_s$ are distinct primes, show that an abelian group of order $p_1p_2...p_s$ must be cyclic."

Now, to show this I want to find an element with order $p_1p_2...p_n$. By Cauchy's theorem, the group contains elements $g_1, g_2$ ..., of order $p_1$, $p_2$, etc. and it is clear that the product of these elements has an order that must be a divisor of $p_1p_2...p_n$. However, I am not sure how to show that the order of this element actually is equal to $p_1p_2...p_n$; i.e., how to make sure there are no cancellations before that?

$\endgroup$
1
  • $\begingroup$ Take a look at the methods used here and generalize them to this case. $\endgroup$
    – superckl
    Oct 30, 2015 at 17:59

1 Answer 1

4
$\begingroup$

We use the following lemma:

Lemma: Let $m$ and $n$ be relatively prime, and suppose $a$ has order $m$ and $b$ has order $n$. Then $ab$ has order $mn$.

Proof: Let $k$ be the order of $ab$. It is clear that $k\le mn$. We have $$b^{km}=(a^{km})(b^{km})=(ab)^{km}=e,$$ and therefore $n$ divides $km$. But $m$ and $n$ are relatively prime, so $n$ divides $k$.

Similarly, $m$ divides $k$, and since $m$ and $n$ are relatively prime, it follows that $mn$ divides $k$, and therefore $k=mn$.

Remark: One can use the same argument to show directly that if $g_i$ has order $p_i$, $i=1$ to $t$, then $g_1g_2\cdots g_t$ has order $p_1p_2\cdots p_t$. But I prefer to go through the Lemma, and then, in principle, induction.

$\endgroup$
1

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.