Consider the group $\mathbb{Z}_{20}$

Question

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.

Answer 1

Any homomorphic image of a cyclic group is cyclic. So you can get that the quotient is cyclic by considering the canonical projection.

The elements of the quotient are the equivalence classes of elements of $\Bbb Z_{20}$ that differ by multiples of $4$.

Since there are five elements in $\langle 4\rangle$, the quotient has order $4$.

Answer 2

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.

Answer 3

One way is to simply list the elements and do it from scratch:

$\mathbb Z_{20}=\{[0]=(0,20,\cdots),\ [1]=(1,21,\cdots),\cdots, [18]=(18,38,\cdots \ ),\ [19]=(19,39,\cdots \ )\}.$

And $[4]=(4,24,\cdots).$ Now, $\mathbb Z$ is Abelian, so the group operation is coset addition.

Then, since $[4]+[4]+[4]+[4]+[4]=[0]$ so $\langle [4]\rangle$ cyclic of order $5$.