How can I prove that if $G$ is a finite group , and the order of $G$ is $pq$ while $p$ and $q$ are primes, and in addition , $G$ with two normalic subgroups , so --> $G$ is cycle? Ideas? Hwo can i show that have just one subgroup for any divisior of pq?
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1what theorems do you have at your disposal? – Gregory Grant Jul 25 '16 at 18:18
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By Sylow theory the two subgroups are then the only non-trivial subgroups of $G$.
Hint: Show that the must exist an element that does not belong to either of those subgroups. What is the subgroup it generates?
Jyrki Lahtonen
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Making my answer CW because I have used this idea at least twice already. Here and here. – Jyrki Lahtonen Jul 23 '16 at 21:01
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How can i do it by this way: to assume there is anothersubgroup of order p, go through "Quotient group" and reach a contradiction – Math_Abst Jul 23 '16 at 21:45
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Let $P, Q$ the subgroups with order $p, q$. As @Jyrki indicated earlier these are the only nontrivial subgroups of G. Let $a \in P, b \in Q$ have orders $p, q$. Consider the element $ab$. Since $p, q$ are relatively prime $a$ is not a power of $b$ and vice versa. Then $ab$ is not in $P, Q$. Since $ab$ is an element of $G$, the order of $ab$ must divide the order of $G$. Then the order of $ab$ is in $\{1, p, q, pq\}$. As $a, b$ are in different groups they are not inverses so it is not $1$. $ab$ is not in $P$ or $Q$ so $ab$ does not have order $p$ or $q$ and therefore must have order $pq$. Then $G$ is generated by $ab$. Then $G$ is cyclic.
Hwai-Ray Tung
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Edited answer. Since $ab \in G$ its order must divide the order of $G$, which makes it either $1, p, q, pq$. More details in edited answer. – Hwai-Ray Tung Jul 23 '16 at 21:29
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You can explain why these are the only nontrivial subgroups of G? and why there must be normal subgroups? it's not clear from his answer – Math_Abst Jul 23 '16 at 21:30
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As @Jyrki stated it's a result of the Sylow theorems. https://en.wikipedia.org/wiki/Sylow_theorems#Theorems It may be better to open another question/check old questions if you would like the proof of those. IIRC the proof of it is somewhat lengthy – Hwai-Ray Tung Jul 23 '16 at 21:34
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Hwo can i get contradict if i assume that has more subgroup with order of p? (it's need be with Quotient group) – Math_Abst Jul 23 '16 at 21:38
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I don't think you need quotient groups or that $P, Q$ are normal for this proof. – Hwai-Ray Tung Jul 23 '16 at 21:45
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How can i do it by this way: to assume there is another subgroup of order p, go through "Quotient group" and reach a contradiction – Math_Abst Jul 23 '16 at 21:46