Prove that $ (\mathbb{Z}/n\mathbb{Z}) / G$ is isomorph to $\mathbb{Z}/ (n/d) \mathbb{Z}$ with $G$ a sub group of $\mathbb{Z}/n\mathbb{Z}$

Question

Question:

Let $d$ a divisor of $n$ ($n,d \in \mathbb{N}$) and let $G$ the only sub group with cardinal $d$ of $\mathbb{Z}/n\mathbb{Z}$. Prove that $ (\mathbb{Z}/n\mathbb{Z}) / G$ is isomorphic to $\mathbb{Z}/ (n/d) \mathbb{Z}$.

-To get a better understanding of how looks like all this sub group you can look here Simple question on a sub group $G$ of $\mathbb{Z}/n\mathbb{Z}$. (basically I needed to understand this in order solve this question :-) )

My answer:

As $G$ is a normal sub group of $(\mathbb{Z}/n\mathbb{Z})$ (trivial) we have by the Lagrange's theorem that $card (\mathbb{Z}/n\mathbb{Z}) / G) = \frac{card( (\mathbb{Z}/n\mathbb{Z}))}{card(G)}= \frac{n}{d}$. More over we have too $card(\mathbb{Z}/ (n/d) \mathbb{Z})=\frac{n}{d}$.
So $(\mathbb{Z}/n\mathbb{Z}) / G)$ and $ \mathbb{Z}/ (n/d) \mathbb{Z}$ have the same finite number of elements.
Now by recursion it is easy to construct a bijection between $(\mathbb{Z}/n\mathbb{Z}) / G)$ and $\mathbb{Z}/ (n/d) \mathbb{Z}$ but I have a problem in order to find a bijectio nthat will be a morphism too (in order to get the asked isomorphism).

Any help (details answer) will be appreciated. Even a lot of different morphism if you succeed to find more than one
Thank you.

Answer 1

I hope that my answer is correct.

In order to answer this question, we are going to build a such isomorphism.

1-
As I explained here all the elements in $ (\mathbb{Z}/n \mathbb{Z} ) / G$ are of the form $a + G$ with $a \in \mathbb{Z}/n \mathbb{Z}$ (that means $ 0 \leq a \in \mathbb{N} \leq n-1$ ).
Now let define $f(a+G)= [a]_{\frac{n}{d}} $ with $[a]_{\frac{n}{d}}$ the equivalent class of $a$ in $ \mathbb{Z}/(n/d) \mathbb{Z}$. That means that $[a]_{\frac{n}{d}}$ is the set of all integers that differ from $a$ by a multiple of $n/d$.
Exemple: $n=8,d=4 \Rightarrow f([3]+G)=[3]_2=[1]_2 \in \mathbb{Z}/2 \mathbb{Z} $

2-
Now let's check that the morphism property is respected.
$f(a+G+b+G)=f(a+b+G)=[a+b]_{\frac{n}{d}}=[a]_{\frac{n}{d}}+[b]_{\frac{n}{d}}=f(a+G)+f(b+G)$
(Edited after the remark of @otr)->For $f$ to be well-defined, it must map equivalent cosets to the same element. If $a + G = b + G$ in $(\mathbb{Z}/n\mathbb{Z}) / G$, then $a - b \in G$. Since $G$ consists of multiples of $n/d$, $a \equiv b \mod (n/d)$. Hence, $f$ is well-defined. See further explanation in part "3-" on injectivity. <-

3-
Surjectivity: by definition of $f$ every element in $\mathbb{Z}/(n/d) \mathbb{Z}$ as a pre image in $ (\mathbb{Z}/n \mathbb{Z} ) / G $. For exemple $ \forall [a] \in \mathbb{Z}/(n/d) \mathbb{Z} \Rightarrow $ we have $a+G$ that gives $f(a+G)=[a]_{\frac{n}{d}} $
Injectivity: By definition if $f(a+G)=f(b+G) \Rightarrow [a]_{\frac{n}{d}}=[b]_{\frac{n}{d}}$ so there $a$ and $b$ have the same euclidian rest after being divised by $n/d \Rightarrow a= \alpha \cdot \frac{n}{d}+r , b = \beta \cdot \frac{n}{d}+r \Rightarrow a-b = (\alpha - \beta) \cdot \frac{n}{d} \in G$ as all the element of $G$ are of the form: $k \cdot \frac{n}{d}$.
And because $a+G=b+G$ iff $a-b \in G$ we get that $a+G=b+G$.

Q.E.D.

Answer 2

Knowing that $card((\mathbb{Z}/n\mathbb{Z})/G) = n/d$ is definitely helpful. What would also be helpful is showing that the quotient of any cyclic group is also cyclic (i.e. $(\mathbb{Z}/n\mathbb{Z})/G$ is cyclic). Now, what are some good candidates $a+G \in (\mathbb{Z}/n\mathbb{Z})/G$ that would generate this entire quotient group?