Let be a cyclic group, an arbitrary group, and and homomorphisms such that and are conjugate subgroups in . If is infinite, assume that and are injective. Prove that by constructing an explicit isomorphism. [Hint: Suppose . Show that for some , for all . Show that the map given by is a homomorphism. Show that is bijective by constructing a two-sided inverse.]
Now , so that for some ; since is cyclic, we have for some . Now let be arbitrary in . Then . Since is arbitrary, we have for all .
We now show that is a homomorphism.
Let . Then
Thus is a homomorphism.
We now show that is bijective.
- Suppose is infinite, so that and are injective. Just as for all , there exists such that for all . Combining these results, we see that . Since is injective, , and since is infinite and arbitrary, we have . Thus .
Define by . Then , so that . Similarly, . Thus is bijective, and we have .
- Before proceeding in the finite case, we prove the following lemma, due to Luís Finotti.
Lemma: Let such that and . Then there exists such that mod and . Proof: Let and write . Now , so that ; we also write and . Let be the product of all prime divisors of which do not divide . Finally, let . Suppose is a prime divisor of . There are three cases:
- If , then since and are relatively prime. Thus does not divide .
- If and , we have two cases.
- If , then by definition. Thus . Also, since , we have . Thus .
- If , then and . Thus .
- If , then since , . Now and , so .
Since no prime divisor of divides , .
Now to the main result.
Suppose is finite. Now is cyclic of order where , by Lagrange. Since generates , generates . Since conjugation by is an isomorphism , generates . Thus . By the Lemma, there exists such that mod and . Moreover, there exists such that mod .
Define by . This map is clearly a two sided inverse of ; hence .