Prove $Z_{20}/\langle 5\rangle \cong Z_{5}$

Question

Im trying to prove the following:

Prove $Z_{20}/\langle 5\rangle \cong Z_{5}$

However, I'm having difficulty.

Here is my attempt:

Note that $\langle 5\rangle=\{5,10,15,\ldots\}$ and $Z_{20}=\{1,2,3,\ldots,20\}$

Observing that, we can see that the left cosets of $Z_{20}$ on $\langle 5\rangle$ are:

$0+\langle 5\rangle=\langle 5\rangle$

$1+\langle 5\rangle=\{6,11...\}$

etc...

But this way, we get that $6+\langle 5\rangle\subset1+\langle 5\rangle$ and so on for other integers.

Now consider the homomorphism $f(zc)=z^{'}\mod5$ where $z$ is in $Z_{20}$, $c$ is in $\langle 5\rangle$ and $z^{'}$ is in $Z_5$

This clearly perseveres the group structure.

However, I'm not sure this is bijective.

Any help would be appreciated. Sorry for writing the proof like this. I'm mediocre at best at mathjax and i wanted it to be readable.

Answer 1

Your quotient group is $$ (\mathbb{Z}/20\mathbb{Z})/(5\mathbb{Z}/20\mathbb{Z}) $$ and the standard homomorphism theorems yield the answer.

Do you want a “computational” answer?

Let $f\colon G\to G'$ be a surjective group homomorphism and $H'$ a normal subgroup of $G'$. Let $$ H=f^\gets(H')=\{x\in G:f(x)\in H'\} $$ If $\pi\colon G'\to G'/H'$ is the canonical homomorphism, then it is easy to see that $\ker(\pi\circ f)=H$ and the homomorphism theorem tells you that $$ G'/H'=\operatorname{im}(\pi\circ f)\cong G/H $$ Now take $G'=\mathbb{Z}/20\mathbb{Z}$ and $H'=5\mathbb{Z}/20\mathbb{Z}$; for $f\colon\mathbb{Z}\to\mathbb{Z}/20\mathbb{Z}$ we take the canonical homomorphism. Then, clearly, $H=5\mathbb{Z}$ (verify it).

Answer 2

Hint

Take $\varphi :\mathbb Z_{20}\to \mathbb Z_5$ defined by $$\varphi (x+20\mathbb Z)=x+5\mathbb Z.$$

Surjectivity is clear. Compute the kernel, and conclude using the first isomorphism theorem.

Answer 3

Let $x=20q + r$ for some integer $q$ and $0 \leq r < 20$, so a representative of the coset containing $r$ in $Z_{20}$. We can then write $x=20q + 5k + r'$ with $0 \leq r' < 5$ and use that $20q+5k=5(4q+k)$ to conclude that this group is ismorphic to $Z_5$.

Answer 4

The quotient has order $20/4=5$, and the only group of order $5$ is $\mathbb{Z}/5\mathbb{Z}$ (up to isomorphism). But let's suppose we don't know this latter fact. Then, we can pursue at least the following alternative options to get the result:

• $x+\langle5\rangle=y+\langle5\rangle \iff x-y+\langle5\rangle=0 \iff x-y\equiv 0 \pmod 5$, and hence the $5$ distinct cosets of $\langle5\rangle$ (the elements of the quotient) can be represented as: $\langle5\rangle,\space\space 1+\langle5\rangle, \space\space 2+\langle5\rangle, \space\space 3+\langle5\rangle, \space\space 4+\langle5\rangle$. Prove that some of them (any but the unit $\langle5\rangle$, actually), times $2,3,4,5$ (additive group!), gives all the other $4$ elements (definition of cyclic group).
• Prove that the map $\mathbb{Z}/20\mathbb{Z}\to \mathbb{Z}/5\mathbb{Z}$, defined by $[x]_{20}\mapsto [x]_{5}$, is a surjective homomorphism with kernel $\langle5\rangle$; then apply the First Homomorphism Theorem. [ EDIT. As correctly pointed out in a comment of another answer, we need first to prove that such a map is well-defined, as we are defining it by means of a particular choice of the representative ("$x$"). So, pick $x'\in [x]_{20}$; thence, $x'\equiv x \pmod {20}$, which in turn implies $x'\equiv x \pmod {5}$, and finally $x'\in [x]_5$. Therefore, $[x']_{20}\mapsto [x']_5=[x]_5$. ]