Proving the Property (a^-1)^-1 = a in a Group | Homework Question

[Benzoate]

Homework Statement

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

Homework Equations

no equations required

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?

 

[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.

 

[Benzoate]

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.

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?

 

[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 ?

 

[matt grime]

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

 

[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, $$(a^{-1})^{-1} = a$$.

 

[olgranpappy]

This method seems circular.

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.