In any group G prove that
Let be a 1-1 mapping of G onto itself such that and let be a 1-1 mapping of G onto itself such that
then
Is this a valid argument? I'm just getting started with abstract algebra.
In any group G prove that
Let be a 1-1 mapping of G onto itself such that and let be a 1-1 mapping of G onto itself such that
then
Is this a valid argument? I'm just getting started with abstract algebra.
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 $(a^{-1})^{-1}=a$.
Couldn't I prove first that for a 1-1 and onto mapping of G onto itself that are inverses by showing 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.
??
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.