2

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.

Shaun
  • 44,997
Neha Aggarwal
  • 71
  • 7
  • 2
    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 $x^k=\phi(b)\implies \left(\phi^{-1}(x)\right)^k=b$. – lulu Nov 27 '19 at 14:27
  • 1
    Welcome to Maths SX! It simply says that solutions of both equations are in bijection with each other. – Bernard Nov 27 '19 at 14:33
  • But x is not in any of the group so how can we apply $\phi$ or $\phi^{-1}$ on it – Neha Aggarwal Nov 27 '19 at 14:38
  • And how can we use this argument to prove this – Neha Aggarwal Nov 27 '19 at 14:39
  • @NehaAggarwal although symbol "$x$" is used for both equations it should be understood as $x\in G$ in the first one and $x\in G'$ in the second one. – freakish Nov 27 '19 at 14:39
  • I think you are just confused because they use the same symbol, $x$, in each case. If it helps, change one of them to $y$. – lulu Nov 27 '19 at 14:41
  • Thanks @Bernard i got this now...but still can't think of the proof – Neha Aggarwal Nov 27 '19 at 14:46
  • Thanks, i understood this now – Neha Aggarwal Nov 27 '19 at 14:49
  • Related, and may help your intuition. In one sense this theorem does not need its own proof: https://math.stackexchange.com/questions/1549008/what-does-it-mean-when-two-groups-are-isomorphic – Ethan Bolker Nov 27 '19 at 14:49
  • It's really ba proof, just an observation: consider the restriction of $\phi$ to the set of roots of the equation in $G$ . It remains a bijection onto the images of these roots, and use the properties of group morphisms. – Bernard Nov 27 '19 at 14:50
  • @lulu sir, $x^k=\phi(b)\implies \left(\phi^{-1}(x)\right)^k=b$. Then ? – Akash Patalwanshi Feb 10 '20 at 06:14
  • 1
    @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 $G$. – lulu Feb 10 '20 at 11:36

1 Answers1

4

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

$$\phi(a)^k=\phi(a^k)=\phi(b)$$

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

$$q=\phi(\phi^{-1}(q))$$

and

$$\phi(b)=q^k=\phi(\phi^{-1}(q)^k)$$

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 $\phi(A)$. $\Box$

freakish
  • 42,851