Let and be groups, let be a homomorphism, and let . Identify and as subgroups of .

Prove that . (Recall that .)

Let . By definition, if , then . By the definition of multiplication in a semidirect product, we have , so that . Comparing first coordinates, . Since is arbitrary, we have , so that .

Suppose . Then . Now let be arbitrary. Then . Thus .

Advertisements

## Comments

If we identify and with their isomorphic copies in , it will be much simpler: iff iff .