# Inverse of an element in a group

• Aug 15th 2010, 03:43 PM
novice
Inverse of an element in a group
Let $\displaystyle G$ be a group. Then

(a) $\displaystyle G$ has a unique identity, and
(b) each element in $\displaystyle G$ has a unique inverse.

The table below shows an abelian group with $\displaystyle 3$ elements:

$\displaystyle \begin{tabular}{lccr} *&a&b&c\\ \cline{2-4}a&a&b&c\\ b&b&c&c\\ c&c&a&b \end{tabular}$

I can see the identity is show on the first row and column, but I don't see the inverse of each elements. Does it mean that the inverses are not in the group?
• Aug 15th 2010, 03:49 PM
pickslides
Quote:

Originally Posted by novice
(b) each element in $\displaystyle G$ has a unique inverse.

You are correct, each element does not have an inverse, therefore $\displaystyle G$ is not a group.
• Aug 15th 2010, 03:50 PM
pickslides
Quote:

Originally Posted by novice

The table below shows an abelian group with $\displaystyle 3$ elements:

$\displaystyle \begin{tabular}{lccr} *&a&b&c\\ \cline{2-4}a&a&b&c\\ b&b&c&c\\ c&c&a&b \end{tabular}$

Also $\displaystyle bc \neq cb \implies G$ does not commute.
• Aug 15th 2010, 03:56 PM
novice
Let $\displaystyle G$ be group. Would it look like this:

$\displaystyle G=\{e, 1/a, a, 1/b, b,...\}$ provided that $\displaystyle G$ is associative and commutative?
• Aug 15th 2010, 04:21 PM
pickslides
Quote:

Originally Posted by novice
Let $\displaystyle G$

$\displaystyle G=\{e, 1/a, a, 1/b, b,...\}$ provided that $\displaystyle G$ is associative and commutative?

I don't think so as $\displaystyle a=e$ and the inverses $\displaystyle a^{-1},b^{-1}, c^{-1}$ are not listed in the make up. You would need a rule linking the other elements.
• Aug 15th 2010, 04:22 PM
undefined
Quote:

Originally Posted by novice
Let $\displaystyle G$ be group. Would it look like this:

$\displaystyle G=\{e, 1/a, a, 1/b, b,...\}$ provided that $\displaystyle G$ is associative and commutative?

A group (G, *) is always associative by definition, but not always commutative. It is commutative if and only if it is abelian, by definition.

In general the inverse of $\displaystyle \displaystyle a$ is denoted $\displaystyle \displaystyle a^{-1}$. Also keep in mind that it's possible to have $\displaystyle \displaystyle a=a^{-1}$, so your enumeration of the set above could have duplicates. There would also be duplicates if you listed a pair of inverses twice, for example if $\displaystyle \displaystyle a = b^{-1}$.
• Aug 15th 2010, 04:33 PM
novice
I see. No wonder $\displaystyle a*b=b$.
• Aug 15th 2010, 04:39 PM
novice
Quote:

Originally Posted by undefined
In general the inverse of $\displaystyle \displaystyle a$ is denoted $\displaystyle \displaystyle a^{-1}$. Also keep in mind that it's possible to have $\displaystyle \displaystyle a=a^{-1}$, so your enumeration of the set above could have duplicates. There would also be duplicates if you listed a pair of inverses twice, for=e, where e is an indentity of (G,*) example if $\displaystyle \displaystyle a = b^{-1}$.

So if $\displaystyle G$ is a group, the inverse of each element is in $\displaystyle (G,*)$. Yah?
• Aug 15th 2010, 04:57 PM
Defunkt
Yes, by definition of a group.
• Aug 16th 2010, 12:13 AM
Swlabr
Quote:

Originally Posted by novice
Let $\displaystyle G$ be a group. Then

(a) $\displaystyle G$ has a unique identity, and
(b) each element in $\displaystyle G$ has a unique inverse.

The table below shows an abelian group with $\displaystyle 3$ elements:

$\displaystyle \begin{tabular}{lccr} *&a&b&c\\ \cline{2-4}a&a&b&c\\ b&b&c&c\\ c&c&a&b \end{tabular}$

I can see the identity is show on the first row and column, but I don't see the inverse of each elements. Does it mean that the inverses are not in the group?

A quick check to see that this is not a group is to note that in each row and column of the Cayley table every element must occur. This reflects the fact that one can get from each element of a group to every other, for example one can get to $\displaystyle c$ from $\displaystyle a$ by (post-)multiplying $\displaystyle a$ by $\displaystyle a^{-1}c$.