Let G be a group. Prove that |Inn(G)|=1 if and only if G is abelian Solution con

ID: 3076010 • Letter: L

Question

Let G be a group. Prove that |Inn(G)|=1 if and only if G is abelian

Explanation / Answer

consider the map f:G-->Inn(G) given by g--->(x--->gxg^-1). f(gh) is the map x-->(gh)x(gh)^-1 = g(hxh^-1)g^-1 = f(g)f(h), so f is a homomorphism of G onto Inn(G). now if g is in ker(f), then f(g) is the identity map, so gxg^-1 = x for all x in G. thus gx = xg, so g is in the center of G, Z(G). thus G/Z(G) is isomorphic to Inn(G). if G is abelian, Z(G) = G, so Inn(G) ˜ G/G = {e} (Inn(G) is trivial). on the other hand, if Inn(G) is trivial, Z(G) must equal G, and G is abelian.

Navigate

Let G be a group, p a prime, and H = {x epsilon G|o(x) = ps, some s 0}. If H Sol

Let G be a loop free graph connected graph, Prove if G is planar, then there exi

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.

Let G be a group. Prove that |Inn(G)|=1 if and only if G is abelian Solution con

Question

Explanation / Answer

Related Questions

Navigate