If $\phi : G \rightarrow G'$ is an isomorphism, then to prove that for a fixed integer $k$ and a fixed group element $b \in G$, the equation $x^k=b\:$ has the same number of solutions in $G$ as does the equation $x^k= \phi(b)$ in $G'$.

I can't visualize what is this theorem trying to say and also looking for a proof.

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    $\begingroup$ I assume that ought to read $x^k=\phi(b)$. Assuming that: just argue that $x^k=b\implies (\phi(x))^k=\phi(b)$ and $x^k=\phi(b)\implies \left(\phi^{-1}(x)\right)^k=b$. $\endgroup$
    – lulu
    Nov 27, 2019 at 14:27
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    $\begingroup$ Welcome to Maths SX! It simply says that solutions of both equations are in bijection with each other. $\endgroup$
    – Bernard
    Nov 27, 2019 at 14:33
  • $\begingroup$ But x is not in any of the group so how can we apply $\phi$ or $\phi^{-1}$ on it $\endgroup$ Nov 27, 2019 at 14:38
  • $\begingroup$ And how can we use this argument to prove this $\endgroup$ Nov 27, 2019 at 14:39
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    $\begingroup$ @AkashPatalwanshi What I wrote establishes a bijection between the solutions. Namely, the map given by $x\mapsto \left(\phi^{-1}(x)\right)$ is a bijection from the relevant solutions in $G'$ to the relevant solutions in $G$. $\endgroup$
    – lulu
    Feb 10, 2020 at 11:36

1 Answer 1


This theorem is simply saying that if two groups are isomorphic then corresponding equations have the same number of solutions. Intuitively, algebra over one group is the same as algebra over the other.

As for the proof: let $A$ be the set of all solutions to $x^k=b$. Since $\phi$ is a homomorphism then for any $a\in A$ we have


meaning $\phi(a)$ is a solution to $x^k=\phi(b)$. And therefore the image $\phi(A)$ is a subset of all solutions to $x^k=\phi(b)$.

On the other hand if $q\in G'$ is a solution to $x^k=\phi(b)$ then since $\phi$ is an isomorphism we have




meaning $b=\phi^{-1}(q')^k$ and thus $q\in\phi(A)$. All in all $\phi(A)$ is equal to the set of all solutions to $x^k=\phi(b)$. Finally since $\phi$ is an isomorphism then $A$ is equinumerous with $\phi(A)$. $\Box$

  • $\begingroup$ Thank you very much $\endgroup$ Oct 12, 2021 at 4:25

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