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Let $G$ be a group and $A=\{g\in G\mid\langle g,x\rangle\text{ is cyclic for all }x\in G\}$.

Why is $A$ a cyclic subgroup of $G$? (Must $G$ be finite?)

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    $\begingroup$ what does it mean "cyclic of $G$" ? Do you mean $A$ is cyclic ? $\endgroup$ – mesel Jun 11 '15 at 6:42
  • $\begingroup$ @mesel. Yes . At first Why $A$ subgroup and second why is Cyclic. $\endgroup$ – A.G Jun 11 '15 at 7:06
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    $\begingroup$ $G$ must be finite; if $G=\mathbb{Q}$, then $A=G$ is not cyclic. If $G$ is finite, consider an element of $A$ of largest order. $\endgroup$ – Slade Jun 11 '15 at 7:06
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Indeed $G$ must be finite, because if $p$ is prime :

$$G=\{z\in S^1| \exists k\geq 0 \text{ with } z^{p^k}=1\} $$

Here $G$ is the set of roots whose order is some power of $p$. One can verify that here $A=G$ (every proper subgroup of $G$ is cyclic). However $G$ is not cyclic (not even generated by one element). Now if $G$ is supposed to be finite then $A$ must be finite as well. $A$ contains $1_G$ and is stable by inverse because $\langle a,x\rangle=\langle a^{-1},x\rangle$. Take now $a,b\in A$, a first remark is that $A\subseteq Z(G)$ because for any $x\in G$, $\langle a,x\rangle$ is cyclic. Now we have for all $x\in G$, $\langle b,x\rangle$ is cyclic generated by $c$ because $b\in B$. Hence : $\langle ab,x\rangle\subseteq \langle a, \langle b,x\rangle\rangle=\langle a,c\rangle$ which is cyclic because $a\in A$, from this we get that $\langle ab,x\rangle$ is a subgroup of a cyclic group hence it is cyclic and you are done showing that $A$ is a subgroup.

Set :

$$A:=\{a_1,...,a_r\} $$

Then one sees that $\langle a_1,a_2\rangle$ is cyclic generated by $b_1\in A$, we see that $\langle b_1\rangle$ contains $a_1$ and $a_2$ then $\langle b_1,a_3\rangle$ is cyclic generated by $b_2$ (we see that $\langle b_2\rangle$ contains $a_1,a_2,a_3$)... and so on $\langle b_{r-2},a_r\rangle$ is cyclic generated by $b_{r-1}$. Now the subgroup generated by $b_{r-1}$ contains all elements $a_1...a_r$ hence contains $A$...

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  • $\begingroup$ Why $b_{1}\in A$? Why $A$ is a subgroup? $\endgroup$ – A.G Jun 11 '15 at 7:48
  • $\begingroup$ You have that $b_1\in <a_1,a_2>\subseteq A$ @A.G. $\endgroup$ – Clément Guérin Jun 11 '15 at 7:56
  • $\begingroup$ Why $\langle a_{1}, a_{2}\rangle\subset A$? At first you must prove that $A$ is subgroup. $\endgroup$ – A.G Jun 11 '15 at 7:59
  • $\begingroup$ @A.G. you are right, clearly there is a lack of argument here. I will edit directly my answer. $\endgroup$ – Clément Guérin Jun 11 '15 at 8:06
  • $\begingroup$ @A.G. I have added the argument to show that $<ab,x>$ is cyclic the idea is just to show that because $<b,x>=<c>$ you have that $<ab,x>\subseteq <a,c>$ and then reapply the $A$-property to have that $<a,c>$ is cyclic. Then $<ab,x>$ is a subgroup of a cyclic group and you are done... $\endgroup$ – Clément Guérin Jun 11 '15 at 8:16

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