Let (G1,∘) and (G2,⋅) be two groups and ϕ:G1⇒G2 be an isomorphism. a) G2 might not be abelian even if G1 is abelian. b) G2 is abelian if and only if G1 is cyclic c) G2 is finite if G1 is finite d) G2 might be abelian even if G1 is abelian

Question

Let (G1,∘) and (G2,⋅) be two groups and ϕ:G1⇒G2 be an isomorphism.  a) G2 might not be abelian even if G1 is abelian.  b) G2 is abelian if and only if G1 is cyclic  c) G2 is finite if G1 is finite  d) G2 might be abelian even if G1 is abelian

mixtyggc · Accepted Answer

Solution: G1andG2 two groups    ϕ:G1→G2 be an isomorphism    ⇒f(g1,g2)=f(g1)⋅f(g2)∀g1g2∈G1,    ∴G1, is abelian    g1g2=g2g1∀g1,g2∈G1,    f(g1,g)P2)=f(g2g1)    (∴f is 1−1)    f(g1)f(g2)=f(g2)f(g1)∀f(g1),f(g2)∈G2    ⇒G2 is abelian group    option (D) correct

Let (G_{1}, \circ)\ \text{and}\ (G_{2}, *) be two groups and \phi:G_{1} \Rightarrow G_{2}ZS

Answered question

Answer & Explanation

New Questions in Abstract algebra