For that reason $gH=(aH)^ i$ for any kind of coset $gH$. so $G/H$ is cyclic, necessarily of cyclic teams.

Exactly how $gH=(aH)^ i$ of any type of coset $gH$.

verifies variable team to be cyclic.

Please clarify.



The debate reveals that each $gH \ in G/H$ is a power of the solitary set component $aH$. Simply put, $G/H =\ (aH)^ i: i \ in \ Bbb Z \ =\ langle aH \ rangle$: the cyclic subgroup of $G/H$ produced by $aH$ is the entire team $G/H$, which is for that reason cyclic.


Below is a somewhat various evidence that I really hope will certainly clear up points. A team $G$ is cyclic if, and also just if, there is a surjective homomorphism $\ pilsadiet.combb Z \ to G$. Currently, think about any type of element team $G/H$. Then there is the approved surjection $G \ to G/H$. Currently, if $G$ is cyclic then there is a surjective homomorphism $\ pilsadiet.combb Z \ to G$. The composite $\ pilsadiet.combb Z \ to G \ to G/H$ is then a surjective homomorphism (given that the compound of surjections is a surjection), therefore $G/H$ is cyclic.


I simply wished to discuss that even more normally, if $G$ is produced by $n$ aspects, then every variable team of $G$ is created by at a lot of $n$ aspects:

Allow $G$ be produced by $\ $, as well as allow $N$ be a regular subgroup of $G$. Then every coset of $N$ in $G$ can be shared as an item of the cosets $Nx_1, \ ldots, Nx_n$. So the established $\ $ creates $G/N$, as well as this collection consists of at many $n$ components.

(Note that the cosets $Nx_i$ will certainly not all stand out if $N$ is non-trivial, however it"s great to create the collection in this manner, equally as $\ $ is a flawlessly legitimate summary of the collection of non-negative actual numbers.)

The outcome concerning cyclic teams is then simply the grandfather clause $n=1$ of this.


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