Let us consider Abelian group "G" and its identity mapping "\\varphi: G\\to G,\\ \\varphi(x)=x." If "x_1\\ne x_2," then "\\varphi(x_1)=x_1\\ne x_2=\\varphi(x_2)," and hence the mapping is one-to-one. For any "y\\in G" we have that "\\varphi(y)=y," and consequently, "\\varphi" is onto. Since "\\varphi(xy)=xy=\\varphi(x)\\varphi(y)" for any "x,y\\in G," we conclude that the mapping is a homomorphism.
Answer: d
Comments
Leave a comment