Math Problem Statement

Let G_1 and G_2 be cyclic groups, both of finite order k. And define f : G_1 \rightarrow G_2 by f(a^n) = b^n, where a \in G_1 and b \in G_2, show that this map is surjective

Solution

To show that the map f:G1G2f : G_1 \to G_2, defined by f(an)=bnf(a^n) = b^n (where aG1a \in G_1 and bG2b \in G_2), is surjective, we must prove that for every element yG2y \in G_2, there exists an xG1x \in G_1 such that f(x)=yf(x) = y.

Step-by-step proof:

  1. Cyclic nature of G1G_1 and G2G_2:

    • Since G1G_1 and G2G_2 are cyclic groups of finite order kk, their elements can be written as: G1={e1,a,a2,,ak1},G2={e2,b,b2,,bk1},G_1 = \{ e_1, a, a^2, \dots, a^{k-1} \}, \quad G_2 = \{ e_2, b, b^2, \dots, b^{k-1} \}, where aa and bb are generators of G1G_1 and G2G_2, respectively, and e1e_1, e2e_2 are their respective identities.
    • The order of aa and bb is kk, so ak=e1a^k = e_1 and bk=e2b^k = e_2.
  2. Definition of the map ff:

    • The map ff is defined as f(an)=bnf(a^n) = b^n, where nn is an integer modulo kk (because of the cyclic nature).
  3. Surjectivity of ff:

    • To show ff is surjective, we need to demonstrate that every element in G2G_2 is in the image of ff. In other words, for any bmG2b^m \in G_2, we need to find an anG1a^n \in G_1 such that f(an)=bmf(a^n) = b^m.
  4. Existence of a preimage:

    • Since f(an)=bnf(a^n) = b^n, for any bmG2b^m \in G_2, we choose ana^n in G1G_1 such that n=mmodkn = m \mod k. This is always possible because nn ranges over all integers modulo kk as ana^n runs through G1G_1.
    • Thus, for any bmG2b^m \in G_2, we have f(am)=bmf(a^m) = b^m.
  5. Conclusion:

    • Every element of G2G_2 has a corresponding preimage in G1G_1 under ff. Therefore, ff is surjective.

Additional Notes:

  • The key property used here is that both G1G_1 and G2G_2 are cyclic groups of the same order kk, so the exponents nn and mm can always match modulo kk.
  • Since ff is defined in terms of the exponents, and the cyclic groups have the same size, ff maps all elements of G1G_1 onto G2G_2 without omission.

Would you like a detailed example or further clarification? 😊


Related Questions:

  1. What additional properties can we prove about ff? Is it injective?
  2. How does the surjectivity depend on the groups G1G_1 and G2G_2 having the same order?
  3. Can this result be generalized to non-cyclic groups?
  4. How do the properties of generators aa and bb influence the map ff?
  5. What happens if G1G_1 and G2G_2 have different orders?

Tip:

Always verify the properties of the group (cyclicity, order, etc.) before analyzing a map, as these heavily influence the results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Group Theory
Cyclic Groups
Surjectivity of Functions
Homomorphisms

Formulas

f(a^n) = b^n

Theorems

Surjectivity of Group Homomorphisms
Properties of Cyclic Groups

Suitable Grade Level

Undergraduate Level