6
$\begingroup$

If$ |G|=pq $ and $p $ doesnt divide $(q-1)$ and $p <q$ then $G$ is cyclic.

My proof stands as this .I have used the fact that

$(i)$ If $|G|=pq$ then I showed that there will only be one element of order $p$ and one element of order $q$.

My approach in proving this part has been to show that if there are two elements of order $p$ say $x_1$ and $x_2$ , then say $H_1$ is a group of order $p$ generated by $x_1$ and $H_2$ is a group of order $p$ generated by $x_2$ .We assume that the intersection is {e} if not then we can get $H_1$=$H_2$(by property of subgroup).

Proving that the elements are distinct .

We assume that the elements $(x_1)^{i}.(x_2)^j$ are not distinct then $(x_1)^{i}.(x_2)^j =(x_1)^{i'}(x_2)^{j'}$.From here we can arrive at a contradiction as $H_1 \cap H_2 =e$.So if there are $p^2$ elements then we can arrive at a contradiction as $p$ and $q$ are both primes.

Similar results will hold in the case of $q$.

$(ii)$ Now there is only one subgroup of order $p$ and one subgroup of order $q$ so they are both normal

$(iii)$ let $H$ and $K$ be two subgroups of order $p$ and order $q$.Then we know that $H \cap K={e}$.$H$ and $K$ are both normal .Then I showed that $x^{-1}y^{-1}xy \in H \cap K$ and $xy=yx$. So the order of the element $xy$ is $pq$.Where am I going wrong in my proof and since I have not used the fact that $p $ doesnot $q-1$.

$\endgroup$
8
  • 2
    $\begingroup$ You use the fact that $p \nmid q -1$ when you prove the uniqueness of the subgroups of order $p, q$ initially. You have to make use of the Third Sylow Theorem. $\endgroup$
    – Colver
    Nov 25, 2020 at 11:34
  • $\begingroup$ While proving that $H_1$ is the only subgroup of order $p$ did I use it ? I tried doing the proof without using sylows theorem.could you just tell me if my proof in the part where I am trying to show that there is only subgroup of order p is correct or wrong? $\endgroup$
    – Antimony
    Nov 25, 2020 at 11:48
  • $\begingroup$ In cases like this it my be most instructive to look at a small example where your argument fails. Consider the case $p=2, q=3$ and the group $S_3$ of order $2\cdot3$. It has three subgroups of order two that I am sure you can describe. What happens for example when $x_1=(12)$ and $x_2=(13)$? The elements are $x_1^ix_2^j$ are distinct all right. There is room for these $4$ elements. But what is the contradiction with there being four distinct elements in a group of order six? $\endgroup$ Nov 25, 2020 at 12:08
  • $\begingroup$ Yes right so is there a better way to use show that there is only one subgroup of order p without using sylows theorem or centre of the group? $\endgroup$
    – Antimony
    Nov 25, 2020 at 12:14
  • 2
    $\begingroup$ $p \lt q$ should be added. $\endgroup$ Nov 25, 2020 at 13:11

4 Answers 4

3
$\begingroup$

An argument close to the OP's idea could proceed as follows. Fill in the details.

  1. If $x$ and $y$ are elements of order $q$, then among the products $x^iy^j$, $0\le i,j<q$, there must be repetitions. This is because $q^2>|G|$. Show that this implies that the subgroup $H$ of order $q$ is unique. Let's fix a generator $x$ of $H$.
  2. If $z\notin H$ has order $pq$ then $G$ is cyclic. Therefore the remaining possibility is that all such elements $z$ have order $p$.
  3. Because $H$ is a unique subgroup of its order, $H\unlhd G$. Why does it follow that $zxz^{-1}=x^i$ for some $i, 1\le i<q$?
  4. Why do we have $z^pxz^{-p}=x$?
  5. On the other hand we also have $z^pxz^{-p}=x^{i^p}$, why? Why does this imply the congruence $$i^p\equiv1\pmod q?$$
  6. It follows that the coset of $i$ in the multiplicative group $\Bbb{Z}_q^*$ has either order $1$ or order $p$. Why?
  7. If the order of the coset of $i$ is equal to one, then $zx$ has order $pq$. Why?
  8. If the order of the coset of $i$ is equal to $p$, why does it follow that $p\mid q-1$?
$\endgroup$
2
  • $\begingroup$ Basically using the known structure of the group of automorphisms of $C_q$. $\endgroup$ Nov 25, 2020 at 15:23
  • $\begingroup$ @smita Correct. Or may be just the group $\Bbb{Z}_q^*$. You do know the order of that group from an earlier course, don't you? $\endgroup$ Nov 26, 2020 at 6:34
3
$\begingroup$

As Gerry Meyerson pointed out, it does not hold in general as it is stated. You need to infer that $p \lt q$. Choose $Q$ a subgroup of $G$ of order $q$ (you can use Cauchy's Theorem for its existence!). Then $|G:Q|=p$ is the smallest prime dividing $|G|$ (ah yes here we are using $p \lt q$), hence $Q \lhd G$ (I hope you know this theorem ... see here for example).

Now $P$ acts on $Q$ by conjugation, but since $p \nmid q-1$ and Aut$(Q) \cong C_{q-1}$ (here we use that $q$ is prime), the action must be trivial: $P$ centralizes $Q$ and the other way around. Since $G=PQ$, $G$ must be abelian and the Chinese Remainder Theorem does the rest: $G \cong C_p \times C_q \cong C_{pq}$.

Bonus remark if $|G|=n$ and gcd$(\varphi(n),n)=1$, then $G$ is cyclic (as a matter of fact the only group of order $n$).

$\endgroup$
5
  • $\begingroup$ I appreciate your answer but I think I have mentioned I dont want to use group actions and the 1st theorem you will not be able to prove without group actions... $\endgroup$
    – Antimony
    Nov 25, 2020 at 15:23
  • $\begingroup$ @smita In a sense my answer runs along similar lines. Other than I don't invoke Cauchy nor that theorem :-) I just outline how you can reach essentially the same conclusions. As Nicky explained, $p\nmid q-1$ comes from the order of the group of automorphisms of $C_q$. I tried to write it in more elementary language, but the idea is the same (so +1). $\endgroup$ Nov 25, 2020 at 15:27
  • 1
    $\begingroup$ Yes true, Jyrki provided an answer there, but the point is if you are doing group theory and "blocking" yourself away from some well-known theorems, it will, not always, encompass some difficulties. But do not get me wrong: if you can do it simple(r) and non-conventional, I strongly support that!! As such your question is very good and I upvoted it! And the answer of Jyrki too! $\endgroup$ Nov 25, 2020 at 15:28
  • $\begingroup$ Thanks a lot , I just wanted to see if there's another way to think about the problem . However I find I am making some basic mistakes .. $\endgroup$
    – Antimony
    Nov 25, 2020 at 16:33
  • $\begingroup$ This is not a problem, that is why this site exists: you showed some original thinking of your own and people helped. You got at the time I wrote this 5 upvotes for your post. Excellent job and keep asking more questions like that! $\endgroup$ Nov 26, 2020 at 10:03
1
$\begingroup$

If $|Z(G)|=p$, then the class equation yields: $pq=p+kq$, a contradiction because $q\nmid p$.

If $|Z(G)|=q$, then the class equation yields: $pq=q+lp$, again a contradiction because $p\nmid q$.

If $|Z(G)|=1$, then the class equation yields: $$pq=1+mq+np \tag1$$ So, there are $np$ elements of order $q$, grouped in $n'$ subgroups of order $q$, and then: $$np=n'(q-1) \tag2$$ Since by assumption $p\nmid q-1$, from $(2)$ follows $p\mid n'$, say $n'=n''p$, whence: $$n=n''(q-1) \tag3$$ which plugged in $(1)$ gives: $$pq=1+mq+n''p(q-1)\tag4$$ But, from $(1)$: $$1+mq=m'p \tag5$$ which plugged in $(4)$ yields: $$q=m'+n''(q-1) \tag6$$ which implies $m'=n''=1$ and then, from $(5)$: $$1+mq=p \tag7$$ a contradiction, because $q>p$.

Therefore $Z(G)=G$, namely $G$ is abelian, and hence (Cauchy) cyclic.

$\endgroup$
-1
$\begingroup$

Consider the center $Z(G)$ of G. The center $Z(G)$ is a normal subgroup of $G$. In particular the order of $Z(G)$ divides the order of $G$ by Lagrange's theorem. So the possible orders of $Z(G)$ are $1,p,q $ or $pq$. Now look at the order of the quotient group $G/Z(G)$.

$\endgroup$
4
  • 4
    $\begingroup$ You need to explain how you rule out $Z(G)=\{1\}$. $\endgroup$ Nov 25, 2020 at 13:56
  • $\begingroup$ If $Z(G)=\{1\}$, then this means that $G$ is not an abelian group. Since every cyclic group is abelian, $G$ is non-abelian implies that it is not cyclic. Does that make sense? $\endgroup$
    – eloparti
    Nov 26, 2020 at 9:17
  • 1
    $\begingroup$ Your reasoning makes perfectly sense, but you need to prove that $G$ is cyclic. You cannot assume that beforehand. $\endgroup$ Nov 26, 2020 at 9:59
  • $\begingroup$ Yes, you are right. I need to employ third Sylow theorem. If $P$ and $Q$ are Sylow p-subgroup and Sylow $q$-subgroup of $G$ respectively, then by the given condition in question $P \trianglelefteq G$ and $Q \trianglelefteq G$. Furthermore $P \cap Q =\{1\}$. So $G\cong P \times Q \implies G\cong C_p \times C_q $ which is abelian. So $Z(G)\neq \{1\}.$ $\endgroup$
    – eloparti
    Nov 26, 2020 at 10:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .