# What are the inverses of $1,3,7,9$ in the group $(G,*_{10})$?

9,405

## Solution 1

There is no "division operation" in a group, it only signifies $a/b$=$ab^{-1}$ So you are right to say that $1/3$ is the inverse of 3, but a better notation would be $3^{-1}$. Now, to see that 3 and 7 are inverses in the group of units mod 10, we see that $3\cdot 7=21 \equiv 1\pmod{10}$ and 1 is the identity. We xan do the same to find the inverse of 9.

## Solution 2

Let $(G, \cdot)$ be a group whose identity is $e$. By definition, for an element $a \in G$, any other $b \in G$ is called inverse of $a$ when $a \cdot b = b \cdot a = e$.

You can actually prove that the inverse of an element is unique. Assume that for $a \in G$ there exist two elements $b, c \in G$ such that both $a \cdot b = b \cdot a = e$ and $a \cdot c = c \cdot a = e$ hold. In particular you will have that $b = b \cdot e = b \cdot a \cdot c = e \cdot c = c$. Given this unicity, we can unambiguously represent the inverse of $a$ by $a^{-1}$.

You are right on saying that the inverse elements are $3^{-1}, 7^{-1}$ and $9^{-1}$, given that they satisfy the axiom above. Actually that is what you are looking for to be satisfied when trying to find the inverse of an element, nothing else. Writing $1/3$ is just a shorthand notation for $1 \cdot 3^{-1}$ in the rational (or real or complex) group; of course you can define it in any other group (like the one you have provided) but using it can be confusing.

Note that the definition of the inverse does not prevent the inverse of an element to be itself. In your example, $9$ is its own inverse not because $9/9 = 1$ (in the rational sense we are used to look at this) but because $9 \cdot 9 = 81 \equiv 1 \mod 10$.

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### Johny

Updated on July 23, 2022

### Comments

• Johny 5 months

Consider $G =%$ {$1,3,7,9$}

The group is $(G,*_{10})$

I am told that the inverse of 1 is itself, which is understandable, as $1/1 = 1$, but I am told that the inverse of 3 is 7, the inverse of 7 is three and the inverse of 9 is itself.

Shouldn't it be, respectively, $3^{-1} , 7^{-1} , 9^{-1}$?

• Johny about 6 years
So finding that the result of multiplication between two elements in agroup is 1 implies inversity between these two elements?
• Johny about 6 years
9 * 9 = 81 (mod 10) = 1. I'm still not sure whether or not the fact that the result of the numbers is the identity of the given operator means that the numbers of each other's inverse?
• ml0105 about 6 years
That is correct- $9$ is its own inverse. Also, a notational issue- We note that $9 \cdot 9 \equiv 1 \pmod{10}$, not $81 \pmod{10} = 1$. The mathematical notation is different than what you may have seen in computer programming.
• pjs36 about 6 years
@Johny Yes. In a group $G$, the inverse $g^{-1}$ of an element $g$ is defined so that $gg^{-1} = 1 =g^{-1}g$; if you multiply two things and get the identity element of the group, those two things are the inverses of each other (by definition).