Proof the set with the multiplication is a group

In summary: Since $\bar{a}\cdot\bar{c}=1$, we have $\bar{c}\cdot\bar{a}=1$, which means that $\bar{c}$ is the left side inverse of $\bar{a}$.In summary, we are trying to prove that ${(\Z/n\Z)}^{\times}$ is an abelian group under multiplication on ${(\Z/n\Z)}$. We know that multiplication is closed and associative under this set, and the identity element is defined as $[1]$. We also have the right side inverse by the definition of the set, and using commutativity of multiplication, we can find the left side inverse.
  • #1
cbarker1
Gold Member
MHB
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Dear Everyone,

$\newcommand{\Z}{\mathbb{Z}}$Suppose the set is defined as:
$\begin{equation*}
{(\Z/n\Z)}^{\times}=\left\{\bar{a}\in \Z/n\Z|\ \text{there exists a}\ \bar{c}\in \Z/n\Z\ \text{with}\ \bar{a}\cdot\bar{c}=1\right\}
\end{equation*}$
for $n>1$
I am having some trouble
Proving that ${(\Z/n\Z)}^{\times}$ is an abelian group under multiplication on ${(\Z/n\Z)}$.

My Attempt:
  • WTS: Multiplication on ${(\Z/n\Z)}^{\times}$ is closed.
By exercise 5 in homework 2, if $[a], \in {(\Z/n\Z)}^{\times}$, then $[a]\cdot \in {(\Z/n\Z)}^{\times}$. So we know that multiplication is closed under ${(\Z/n\Z)}^{\times}$.

  • Associativity:
Let $[a],,[c]\in{(\Z/n\Z)}^{\times}$ . We know that the $\Z$ is associative.
$[a]\cdot(\cdot[c])=[a] \cdot [bc]=[abc]=[ab] \cdot [c]=([a]\cdot) \cdot [c]$.
  • Identity
We know that $[1]$ is the identity for ${(\Z/n\Z)}^{\times}$.
$[1]\cdot[x]=[1\cdot x]= [x], \forall [x] \in {(\Z/n\Z)}^{\times}$
$[x] \cdot [1]=[x \cdot 1]=[x], \forall [x] \in {(\Z/n\Z)}^{\times}$
  • inverse
We are given the right side inverse by the definition of the set. We need to show that the left side exist ( here is where I am having troubles).

Thanks,
Cbarker1
 
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  • #2
Hi Cbarker1,

Let's do commutativity first:
$$\forall [a],\in(\mathbb Z/n\mathbb Z)^\times:[a]\cdot=[ab]=[ba]=\cdot[a]$$

Now we can apply commutativity of multiplication to find the left side inverse.
 

FAQ: Proof the set with the multiplication is a group

1. What is a group in mathematics?

A group in mathematics is a set of elements that follow a specific set of rules for combining elements together. These rules include closure, associativity, identity, and inverses.

2. What is the operation of multiplication in a group?

In a group, the operation of multiplication is defined as a binary operation where two elements from the group are combined to produce a third element. This operation must follow the rules of closure, associativity, and identity.

3. How do you prove that a set with multiplication is a group?

To prove that a set with multiplication is a group, you must show that it satisfies the four group axioms: closure, associativity, identity, and inverses. This can be done by showing that the operation is closed, associative, has an identity element, and every element has an inverse within the set.

4. What is the significance of proving that a set with multiplication is a group?

Proving that a set with multiplication is a group is significant because it allows us to apply the properties of groups to the set. This can help us solve problems and make predictions about the behavior of the elements in the set.

5. Can a set with multiplication be a group if it does not satisfy all of the group axioms?

No, a set with multiplication cannot be considered a group if it does not satisfy all of the group axioms. The group axioms are essential for defining a group and without them, the set cannot be considered a group.

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