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Let $n$ be a positive integer and let $n\Bbb{Z}=\{nm\mid m \in\Bbb{Z}\}$. I need to show that $\left< n\Bbb{Z},+ \right>$ is a group. And I need to show that $\left< n\Bbb{Z},+ \right>\cong\left< \Bbb{Z},+ \right>$.

Added: If $n\mathbb Z$ is a subgroup of $\mathbb Z$ then it must be closed under "+". The identity element 0 is an element of the subgroup. And for any $n$ in $n\mathbb Z$, it's inverse $-n$ must be in $n\mathbb Z$...

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@Stahl how do you insert the symbol _ with ~ on top? That's the symbol I need there –  user58315 Mar 28 '13 at 1:48
    
You can use \simeq, but in many places, $\cong$ and $\simeq$ are both used to indicate "are isomorphic." –  Stahl Mar 28 '13 at 1:50

3 Answers 3

What are the properties (or axioms) of a group $\langle G, *\rangle$?

  • We need to show that the binary operation $*$ is closed on the set $G$.
  • We need that the binary operation $*$ is associative on $G$.
  • We need to show that there exists an identity element $e \in G$ such that for every element $g\in G,\;e*g = g* e = g$.
  • We need to show that for each element $g \in g$, there exists an inverse element $g^{-1}$ such that $g *g^{-1} = g^{-1} * g = e$.

Note: You can show a $\langle n\mathbb Z, + \rangle$ is a group by showing it is a subgroup of of $\langle \mathbb Z, + \rangle$, which we know is a group. Any subgroup of a group is necessarily a group!

How do we show a group $H$ is a subgroup of another group $G$?

  • We need to show that $H$ is closed under the group operation of $G$.

  • We need to show that the identity element of $G$ is an element of $H$, (and is thus the identity of $H$).

  • We need to show that for each $h \in H$, $h^{-1} \in H$.

When we've done those things, we conclude $H \leq G$, and is hence a group in its own right.

From comments: We want to SHOW $n\mathbb Z \leq Z$.

  • To show closure, show that for $a, b \in n\mathbb Z, \; a = nx, b = ny$ for some integers $x, y \in\mathbb Z$. Then $a + b = nx + ny = n(x+y) \in n\mathbb Z,\;$ since $\;x, y \in \mathbb Z$ and $\mathbb Z$ is a group, therefore closed under addition, we have $x + y \in \mathbb Z$.
  • For the identity: Note that $0 = n\cdot 0$ and $0 \in \mathbb Z$, so $n\cdot 0 = 0 \in n\mathbb Z$.
  • For inverse elements: if $a \in n\mathbb Z$, then there exists an integer $z \in \mathbb Z$ such that $a = nz$. Now, as you noted below, if $z \in \mathbb Z$, then since $\mathbb Z$ is a group under addition, $-z \in \mathbb Z$. So $n(-z) = -nz \in n\mathbb Z$. Hence the (additive) inverse of each element in $n\mathbb Z$ is also in $n\mathbb Z$.

To show that $\langle \mathbb Z, + \rangle \cong n\langle \mathbb Z, \rangle$, we need to show that there exists an isomorphism $\varphi: \mathbb Z \to n\mathbb Z$ such that $\varphi$ is bijective (one-to-one and onto), and that $\varphi$ satisfies the homomorphism property: For every $$i, j \in \mathbb Z, \varphi(x+y) = \varphi(x) + \varphi(y)$$

Now, can you try the map for $\varphi: \mathbb Z \to n\mathbb Z$ defined by $z \mapsto nz$: $\varphi(z) = nz$ for every $z \in \mathbb Z$! Test to make sure $\varphi$ satisfies the properties of an isomorphism.

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If Z is a subgroup of nZ, then z must be closed under "+". The identity element 0 is an element of Z. And for any n in Z it's inverse -n must be in Z –  user58315 Mar 28 '13 at 2:00
    
Note that showing $n\mathbb Z \cong \mathbb Z$ is equivalent to showing $\mathbb Z \cong n\mathbb Z$ –  amWhy Mar 28 '13 at 2:05
    
We want to SHOW $n\mathbb Z \leq Z$. Almost: to show closure, show that for $a, b \in \mathbb Z, \; na + nb = n(a+b) \in n\mathbb Z,$, easy enough to do! For the identity: Note that $0 = n\cdot 0$ and $0 \in \mathbb Z$, so $0 \in n\mathbb Z$. Now, can you try the map for $\varphi: \mathbb Z \to n\mathbb Z$ defined by $z \mapsto nz$: $\varphi(z) = nz$ for every $z \in \mathbb Z$! –  amWhy Mar 28 '13 at 2:13
    
@amWhy: Nice Amy.+1 –  Babak S. Mar 28 '13 at 3:34
    
@amWhy: What a excellent answer! Wish I could give it a double-thumbs-up! +1 –  Amzoti May 14 '13 at 1:42

Hint: To show something is a group, you need to show that it has an identity element, inverses, and an associative binary operation. Think about how you know $\Bbb{Z}$ is a group (what's the identity? what's the inverse of an arbitrary element? is it associative?) as a start on understanding/proving why $n\Bbb{Z}$ is a group (in fact it's a subgroup of $\Bbb{Z}$).

Hint 2: Consider the map \begin{align*} \phi : n\Bbb{Z}&\to\Bbb{Z}\\ nm&\mapsto m. \end{align*} Can you show that this is a bijection that preserves the group structure? (I.e., you will show $\phi$ is a bijective group homomorphism, AKA an isomorphism.)

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Since every element of $n\mathbb Z$ is an element of $\mathbb Z$, we can do an easier proof that it is a group by showing that it is a subgroup of $\mathbb Z$.

It happens to be true that if $H\subset G$, where $G$ is a group, then $H$ is group if it satisfies the single condition that if $x,y\in H$, then $x\ast y^{-1}\in H$, where $\ast$ is the group operation of $G$ and $y^{-1}$ is the inverse of $y$ in $G$.

In order to find an isomorphism $\varphi:\mathbb Z\to n\mathbb Z$, ask yourself what $\varphi(1)$ should be. Can you go from here?

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