Consider the group $\mathbb{Z}_{20}$ and let $H = \langle [4]\rangle $ be the subgroup generated by $4$. List all the elements of $\mathbb{Z}_{20}/H$ and show that the quotient is cyclic.
I think I just have a lack of understanding of terminology here. I know that $\mathbb{Z}_{20}/H$ just represents the left cosets of $H$, but since $H = \langle [4]\rangle$ is the subgroup generated by $4$, what exactly does this mean?
I understand how to show that the quotient is cyclic too, here is my attempt:
If there is an element $a \in G$ such that $\langle a\rangle = G$, we say that G is a cyclic group. However, I don't think we necessarily need this here. From the first part of this question, if the order is prime, then by lagranges theorem, or rather a theorem in my book we have that $\mathbb{Z}_{20}/H$ is cyclic and we are done.
Best Answer
You can understand a quotient in many ways, the most geometric one (in the case of general groups) would be as a collapsing of the corresponding group in the Cayley Graph, but to return to a more algebraic side : Here, $[4]$ is what is called a normal subgroup, that is, its left cosets are the same as his right ones, so $G/H$ is just the set of cosets of $H$. The cosets of a subgroup $H$ are the equivalence classes defined by $a \sim b \equiv aH = \{ ah | h \in H\} = bH$. When you quotient by this subgroup, you take the set of all those cosets and give it the natural product $gHfH = gfH$ to make it a group. Now intuitively you want to divide your group among the ways your subgroup can be "translated" or moved around your group. Here you want to know how elements of $\mathbb{Z}_{20}$ behave under multiplication by $4$. However, for all $g \in \mathbb{Z}_{20}$ you clearly have $(g + 4)H = gH$ thus you are doing "modulo 4" inside $\mathbb{Z}_{20}$, that is $\mathbb{Z}_{20}/H$ is the set of elements of $\mathbb{Z}_{20}$ that produce different subgroups when multiplied with $[4]$.You should have a good idea of what they are with the precedent remark. Now to show it is cyclic you should prove it is again isomorphic to a cyclic subgroup which will be easy once you have deduced what $\mathbb{Z}_{20}/H$ is.