Prove that the inverse of isomorphism is also an isomorphism

[Anood]

Can you please help me in proving that the inverse of an isomorphism is an isomorphism?

 

[pka]

Re: Isomorphism

Can you please help me in proving that the inverse of an isomorphism is an isomorphism

Yes someone can help, but you must provide much more information.

For example, we need to know the underlining algebraic structure: is it a group, ring, field or vector space? How is the mapping defined?

You must post the exact question!

 

[Anood]

it's a group

i think i should start like this if f:G->k then f^-1:K->G

 

[pka]

Suppose that each of \(\displaystyle S\) and \(\displaystyle T\) is a group and \(\displaystyle \phi :S \mapsto T\) is an isomorphism. Because \(\displaystyle \phi\) is a bijection, one-to-one and onto, its inverse \(\displaystyle \phi^{-1}\) clearly exits and is bijective. Now \(\displaystyle \phi\) preserves to operations defined on the underlining group structure: \(\displaystyle \left( {\forall a\,\& \,b \in S} \right)\left[ {\phi (a \circ b) = \phi (a) * \phi (b)} \right]\) where \(\displaystyle \circ\) is the group operation in \(\displaystyle S\) and \(\displaystyle *\) is the operation in \(\displaystyle T\).

Because of onto \(\displaystyle \left\{ {x,y} \right\} \subseteq T \Rightarrow \;\left( {\exists a\,\& \,b \in S} \right)\left[ {\phi (a) = x\;\& \;\phi (b) = y} \right]\) and \(\displaystyle \left( {\exists c \in S} \right)\left[ {\phi (c) = x * y} \right].\)

Because of being one-to-one we have:
\(\displaystyle \begin{array}{l}
\left[ {\phi (a \circ b) = \phi (a) * \phi (b)} \right] \\
\left[ {\phi (a \circ b) = x * y = \phi (c)} \right] \\
\quad \Rightarrow \quad a \circ b = c \\
\end{array}.\)

Now we put all this together.
\(\displaystyle \begin{array}{rcl}
\phi ^{ - 1} \left( {x * y} \right) & = & \phi ^{ - 1} \left( {\phi (c)} \right) \\
& = & c \\
& = & a \circ b \\
& = & \phi ^{ - 1} \left( x \right) \circ \phi ^{ - 1} \left( y \right) \\
\end{array}.\)

DONE!