1. ## group theory

In any group G prove that $\displaystyle (a^{-1})^{-1} = a$

Let $\displaystyle f$ be a 1-1 mapping of G onto itself such that $\displaystyle f(a) = a^{-1}$ and let $\displaystyle f^{-1}$ be a 1-1 mapping of G onto itself such that $\displaystyle f^{-1} (a^{-1}) = a$

then $\displaystyle (a^{-1})^{-1} \equiv (f(a))^{-1} = f^{-1} \circ f(a) = f^{-1}(a^{-1}) = a$

Is this a valid argument? I'm just getting started with abstract algebra.

2. ## Re: group theory

No, I don't think this is valid. First you have to prove that f is a "good" function. The only way I can see to prove this is to prove that inverses are unique. But if you have this, the desired conclusion $(a^{-1})^{-1}=a$ follows immediately from $a^{-1}a=aa^{-1}=1$; i.e. $a$ satisfies the condition to be the inverse of $a^{-1}$. Secondly, I don't see how to prove f is 1-1 and onto without knowing $(a^{-1})^{-1}=a$.

3. ## Re: group theory

Couldn't I prove first that for a 1-1 and onto mapping of G onto itself that $\displaystyle f , f^{-1}$ are inverses by showing $\displaystyle f^{-1} \circ f (a) = a$ then state by the property of being 1-1 and onto that this is a sound argument for proving that a has an inverse a^-1 such that the inverse of the inverse is the original element a because the mapping can be argued as being the operation * under which the group is closed. saying a*a^-1 = 1 seems valid but doesn't seem general enough because then e = 1 which is not general.

??

4. ## Re: group theory

ie $\displaystyle *$ is defined as $\displaystyle f(a) = b$ where $\displaystyle b = a^{-1}$ and by the existence of inverses $\displaystyle f^{-1}(b) = a$

??

or

$\displaystyle b,c \in G \ni ab=e=ac \Rightarrow b=c \Rightarrow b=c=a^{-1}$ and $\displaystyle (a^{-1})^{-1} = b^{-1} = c^{-1}$

then
$\displaystyle b^{-1} = c^{-1} = a$

therefore $\displaystyle (a^{-1})^{-1} =a$
something more along these lines?

5. ## Re: group theory

In a group, inverses are unique, it's a one line proof:

Suppose $a\ast b = b\ast a = e$ and $a \ast c = c\ast a = e$. Then $b = e\ast b = (c \ast a)\ast b = c\ast (a \ast b) = c \ast e = c$.

Now since inverses are unique, ANY $b \in G$ such that $a^{-1} \ast b = b \ast a^{-1} = e$ must be THE inverse of $a^{-1}$.

Clearly $b = a$ works.