Subgroup-defining function value is characteristic

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Suppose f is a subgroup-defining function, i.e., f associates to every group G a subgroup f(G), with the property that given an isomorphism \sigma:G_1 \to G_2, the image of f(G_1) under \sigma is f(G_2).

Then, for every group G, f(G) is a characteristic subgroup of G.

Related facts

Conclusion about normality

Combined with the fact that characteristic implies normal, this tells us that any subgroup-defining function value is normal.

Applications to specific subgroup-defining functions

Subgroup-defining function Meaning Proof that it is characteristic Proof of stronger properties satisfied for the particular subgroup-defining function
center elements that commute with everything center is characteristic center is bound-word, center is strictly characteristic, center is direct power-closed characteristic
derived subgroup products of commutators derived subgroup is characteristic derived subgroup is verbal, derived subgroup is fully invariant
Frattini subgroup intersection of all maximal subgroups Frattini subgroup is characteristic


Given: A group G, a subgroup-defining function f, an automorphism \sigma of G.

To prove: \sigma(f(G)) = f(G).

Proof: Since \sigma:G \to G is an automorphism, it is in particular an isomorphism. Thus, the definition of subgroup-defining function above, setting G_1 = G_2 = G, gives us what we need to prove.