$F:G\to B$ is an isomorphism between groups implies $F^{-1}$ is an isomorphism

$F$ is an isomorphism from group $G$ onto group $B$. Prove $F^{-1}$ is an isomorphism from $B$ onto $G$.

I do not know which direction to go in.

• $F : G \rightarrow B$ is a bijective homomorphism. It is immediate that $F^{-1} : B \rightarrow G$ is bijective, so you need to show that it is a homomorphism. Start by writing out the definition of $F^{-1}$ being a homomorphism, and try to relate it to $F$ being a homomorphism. – Michael Joyce Nov 5 '13 at 13:11

Since $$F$$ is an isomorphism of $$G$$ onto group $$B$$ we know that $$F^{-1}$$ is bijective that being said we know its $$1-1$$ and onto and all you'd have to do is prove that $$F^{-1}$$ is operation preserving.

Hint: if $x,y\in B$, then $x=F(g)$ and $y=F(h)$ for some $g,h\in G$. Then $$F^{-1}(xy)=F^{-1}(F(g)F(h))=\ldots$$

Excuse me for this rather pedantic answer: an isomorphism is a homomorphism$~f$ that has an inverse homomorphism$~g$, and since the conditions for an inverse are symmetric (namely $f\circ g=1$ and $g\circ f=1$), it is clear that $g$ is an isomorphism as well (with inverse$~f$). There is nothing more to prove.

Now I realise that this may not be the definition of an isomorphism of groups that you have at hand; it might be defined as a bijective homomorphism. That is not the correct definition. It is a valid characterisation of isomorphisms of groups, exactly because one can prove that the inverse bijection then is a homomorphism of groups (and of course being bijective is also necessary to have an inverse homomorphism). But before this is shown, there is no justification to call bijective homomorphisms isomorphisms. The definition of isomorphism is always the above one in terms of homomorphisms; it is valid in all categories.

Now it happens that in many algebraic categories (groups, monoids, fields, vector spaces, rings, modules, you name it) all bijective homomorphisms are isomorphisms. This is because in such categories, all properties that a function$~f$ must satisfy to be a homomorphism are of the form $$\def\Op{\operatorname{Op}} f(\Op(x_1,x_2,...))=\Op(f(x_1),f(x_2),...),$$ where $\Op$ is an algebraic operation in the category. (For groups the relevant operations are multiplication (with $2~$arguments), inverse ($1~$argument) and identity ($0~$arguments), though it turns out requiring it for multiplication alone suffices.) Given that such a requirement holds for $f$ and that $f$ has an inverse function$~f^{-1}$, set $y_i=f(x_i)$ for all arguments, so that $x_i=f^{-1}(y_i)$, and apply $f^{-1}$ to both sides of the equation. This gives after simplification $\Op(f^{-1}(y_1),f^{-1}(y_2),...)=f^{-1}(\Op(y_1,y_2,...))$, so that $f^{-1}$ satisfies the same requirement. This applies in particular for groups, with $\Op(x,y)=xy$.

Because $F$ is a bijection, $F^{-1}$ is a bijection as well. Now we know $$F(a\cdot b)=F(a)\cdot F(b)$$ This means that $$F^{-1}(a\cdot b)=F^{-1}(F^{-1}(F(a\cdot b))) = F^{-1}(F^{-1}(F(a)))\cdot F^{-1}(F^{-1}(F(b))) = F^{-1}(a)\cdot F^{-1}(b)$$ Thus $F^{-1}$ is an isomorphism from $B$ to $G$.

• In the second equality it seems that you already use that $F^{-1}$ is a homomorphism. – Dan Shved Nov 5 '13 at 13:18
• That's what I thought as well. – Christoph Nov 5 '13 at 13:18
• Ah yes I see... Sorry was a sloppy mistake but it renders the whole argument invalid. Thanks for pointing it out! egreg answer is indeed the better way to go! – Slugger Nov 5 '13 at 13:50