Therefore $gH=(aH)^i$ for any kind of coset $gH$.so $G/H$ is cyclic , by meaning of cyclic groups.

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

proves variable group to be cyclic.

Please explain.


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The discussion shows the each $gH\in G/H$ is a strength of the single fixed element $aH$. In other words, $G/H=\(aH)^i:i\in\Bbb Z\=\langle aH\rangle$: the cyclic subgroup the $G/H$ generated by $aH$ is the whole group $G/H$, i beg your pardon is thus cyclic.


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Here is a slightly different proof that ns hope will certainly clarify things. A team $G$ is cyclic if, and also only if, there is a surjective homomorphism $\thedailysplash.tvbb Z \to G$. Now, consider any factor team $G/H$. Climate there is the canonical surjection $G \to G/H$. Now, if $G$ is cyclic climate there is a surjective homomorphism $\thedailysplash.tvbb Z\to G$. The composite $\thedailysplash.tvbb Z\to G\to G/H$ is then a surjective homomorphism (since the composite that surjections is a surjection), thus $G/H$ is cyclic.


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I just wanted to cite that more generally, if $G$ is produced by $n$ elements, climate every element group that $G$ is created by at many $n$ elements:

Let $G$ be generated by $\x_1,\ldots x_n\$, and let $N$ be a normal subgroup that $G$. Then every coset that $N$ in $G$ have the right to be expressed together a product of the cosets $Nx_1,\ldots, Nx_n$. For this reason the set $\Nx_1,\ldots,Nx_n\$ generates $G/N$, and this set contains at many $n$ elements.

(Note the the cosets $Nx_i$ will certainly not every be distinctive if $N$ is non-trivial, yet it"s well to create the set this way, simply as $\x^2 \mid x\in \thedailysplash.tvbbR\$ is a perfectly valid summary of the set of non-negative actual numbers.)

The an outcome about cyclic groups is then just the special instance $n=1$ the this.


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