I'm stuck!

Let $G, H$ and $K$ be three groups. Given $f: G \to H$ and $g: H\to K$ are isomorphisms, prove that the composition $g\circ f: G \to K$ is an isomorphism.

  • 3
    $\begingroup$ Are you given that $f$ and $g$ are isomorphisms? $\endgroup$ Dec 10 '14 at 22:07
  • $\begingroup$ Sorry yes they are $\endgroup$
    – maribov
    Dec 10 '14 at 22:11
  • 2
    $\begingroup$ You have two things then to show: $g\circ f$ is indeed a homomorphism (this follows from $f$ and $g$ being homomorphisms) and that it is bijective (this follows from $f$ and $g$ being bijections). I'll leave the details to you. $\endgroup$ Dec 10 '14 at 22:13
  • $\begingroup$ @Cameron: oog, no. Show that if $f$ and $g$ have inverses then $gf$ has an inverse. $\endgroup$ Dec 11 '14 at 1:58

First, what needs to be done in order to show that the mapping $g\circ f$ is an isomorphism? Well, you can simply make sure it first obeys the multiplicativeness law $(g\circ f)(ab)=(g\circ f)(a)(g\circ f)(b)$. It should also be a bijection.

To see that it's a bijection, note that since the individual maps are isomorphisms (and thus bijections), their composition is necessarily a bijection (there's no algebra here, just set theory).

To show the other condition, start by letting $x,y\in G$ and considering $(g\circ f)(xy)$. From there you are able somehow to end up with the RHS of the homomorphism law.


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