# Finding the structure of a group without using sylows theorem.

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

• 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. Nov 25, 2020 at 11:34
• 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? Nov 25, 2020 at 11:48
• 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? Nov 25, 2020 at 12:08
• 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? Nov 25, 2020 at 12:14
• $p \lt q$ should be added. Nov 25, 2020 at 13:11

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 $$x^iy^j$$, $$0\le i,j, there must be repetitions. This is because $$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?
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 $$p\mid q-1$$?
• Basically using the known structure of the group of automorphisms of $C_q$. Nov 25, 2020 at 15:23
• @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? Nov 26, 2020 at 6:34

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 $$n$$).

• 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... Nov 25, 2020 at 15:23
• @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). Nov 25, 2020 at 15:27
• 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! Nov 25, 2020 at 15:28
• 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 .. Nov 25, 2020 at 16:33
• 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! Nov 26, 2020 at 10:03

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.

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 $$G/Z(G)$$.

• You need to explain how you rule out $Z(G)=\{1\}$. Nov 25, 2020 at 13:56
• 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? Nov 26, 2020 at 9:17
• Your reasoning makes perfectly sense, but you need to prove that $G$ is cyclic. You cannot assume that beforehand. Nov 26, 2020 at 9:59
• 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 $Z(G)\neq \{1\}.$ Nov 26, 2020 at 10:33