# Group question

1. Aug 24, 2007

### Benzoate

1. The problem statement, all variables and given/known data

In a group, prove that (a^-1)^-1

2. Relevant equations

no equations required

3. The attempt at a solution

the inverse of a is 1/a and the inverse of 1/a is a. therefore , (a^-1)^-1 = a for all a.

Also for the property of an exponent, (a^n)^m=a^(m*n)
so , (a^-1)^-1=a^(-1*-1)=a,

There for (a^-1)^-1 = a for all a

Did I use the right methods for proving that (a^-1)^-1 =a?

2. Aug 24, 2007

### dextercioby

No, no, the exponent method is not okay. It has nothing to do with this problem in group theory. The exponent "-1" is just a shorthand notation for the inverse in a group, it's not the exponent one meets when working with numbers, real, complex, etc.

So let's see

Do you agree that

$$\left(a^{-1}\right)^{-1} \cdot a^{-1} = e$$ ??

"e" is the identity in the group.

If so, if you agree, then consider multiplying the whole relation to the right by "a". See what you get.

Last edited: Aug 24, 2007
3. Aug 24, 2007

### Benzoate

This method seems circular. Here are the results of my new proof:

(a^-1)^-1 * a^-1 -e => a^(-1*-1) *a^(-1)=a^0 =1 =e

by multiplying a on both sides I get
a*((a^(-1))^-1 *a^-1) = a*e

a*(a^(-1*-1) *a^-1)=e*a
a*(a*a^(-1))=ea
a's cancel out
(a*a^(-1))=e

since e also equals (a^-1)^-1 *a^-1) , then (a*a^(-1))= (a^-1)^-1 *a^(-1)

therefore

((a^(-1))^-1)=a for all a?

Aren't suppose to multiply e on both sides as opposed to multiplying a?

Last edited: Aug 24, 2007
4. Aug 24, 2007

### dextercioby

Come on

$$\left(a^{-1}\right)^{-1}\cdot\left(a^{-1}\cdot a \right)=e\cdot a$$

I used the associativity of the group operation. Now make the 2 multiplications, one inside the bracket and one in the RHS of the eq. What do u get ?

5. Aug 24, 2007

### matt grime

What is the definition of the inverse of a? It is certainly NOT 1/a.

6. Aug 24, 2007

### Kummer

$$a^{-1}$$ is inverse of $$a$$, i.e. $$aa^{-1} = a^{-1} a = 1$$.
So we see that $$a$$ is inverse of $$a^{-1}$$ which means, $$(a^{-1})^{-1} = a$$.

7. Aug 24, 2007

### olgranpappy

no, it's no circular. He was explaining to you that you must use the definition of the (left) inverse. I.e., for any element $$g^{}$$ of the group, the inverse $$g^{-1}$$ satisfies
$$g^{-1}\cdot g = e$$
where the symbol $$e^{}$$ stands of the identity element of the group.

Maybe the notation $$g^{-1}$$ for the inverse of $$g$$ is too familiar for its own good. You could try a different notation. For example, you could denote the inverse of $$g$$ as
$$\bar g$$. Then, the definition of the inverse says that
$$\bar g \cdot g = e$$

To prove the equality you mentioned, first simply rename the element $$g$$ to $$\bar a$$ (which is an element of the group if $$a$$ is, by definition) to see that
$$\bar \bar a \cdot \bar a = e$$
(which is just using the definition of inverse for $$\bar a$$).

Next, rename the element $$g$$ to $$a$$ to see that
$$\bar a \cdot a = e$$
(which, is just using the definition of inverse for a).

If you multiply the above equation on the left by $$\bar \bar a$$ and the equation two-above on the right by $$a$$ you will see that the LH sides are equal and thus the RH sides are also equal which gives you the equality you desire.