Prove that a factor group of a cyclic group is cyclic.
I didn't understand last two lines of proof ..
Therefore $gH=(aH)^i$ for any coset $gH$. so $G/H$ is cyclic , by definition of cyclic groups.
How $gH=(aH)^i$ of any coset $gH$.
proves factor group to be cyclic.
Please explain.
The argument shows that each $gH\in G/H$ is a power of the single fixed element $aH$. In other words, $G/H=\{(aH)^i:i\in\Bbb Z\}=\langle aH\rangle$: the cyclic subgroup of $G/H$ generated by $aH$ is the whole group $G/H$, which is therefore cyclic.
Here is a slightly different proof that I hope will clarify things. A group $G$ is cyclic if, and only if, there is a surjective homomorphism $\mathbb Z \to G$. Now, consider any factor group $G/H$. Then there is the canonical surjection $G \to G/H$. Now, if $G$ is cyclic then there is a surjective homomorphism $\mathbb Z\to G$. The composite $\mathbb Z\to G\to G/H$ is then a surjective homomorphism (since the composite of surjections is a surjection), thus $G/H$ is cyclic.
I just wanted to mention that more generally, if $G$ is generated by $n$ elements, then every factor group of $G$ is generated by at most $n$ elements:
Let $G$ be generated by $\{x_1,\ldots x_n\}$, and let $N$ be a normal subgroup of $G$. Then every coset of $N$ in $G$ can be expressed as a product of the cosets $Nx_1,\ldots, Nx_n$. So the set $\{Nx_1,\ldots,Nx_n\}$ generates $G/N$, and this set contains at most $n$ elements.
(Note that the cosets $Nx_i$ will not all be distinct if $N$ is non-trivial, but it's fine to write the set this way, just as $\{x^2 \mid x\in \mathbb{R}\}$ is a perfectly valid description of the set of non-negative real numbers.)
The result about cyclic groups is then just the special case $n=1$ of this.
