## If the direct factors are characteristic in a direct product, then the automorphism group of the product is the product of automorphism groups

Prove that if and are groups such that and are characteristic in , then . Deduce that if is an abelian group of finite order then is isomorphic to the direct product of the automorphism groups of its Sylow subgroups.

We define a mapping as follows: . We claim that is an isomorphism.

- ( is a homomorphism) Suppose all mappings are in the appropriate sets. Then .
- ( is injective) Suppose . Then for all and , we have . Since pairs are equal precisely when their corresponding entries are equal, we have and for all and ; thus and , and we have .
- ( is surjective) Let . Let ; since is characteristic, . Define a mapping by . We claim that is an automorphism of .
- ( is a homomorphism) Note that . Comparing entries, we have for all .
- ( is injective) Suppose . Then . Since is injective, .
- ( is surjective) Let . Since , there exists such that . Thus .

Likewise, we can define a mapping which is also an isomorphism.

Finally, note that Thus , and we have .

Thus .

As a lemma, we can extend this result by induction to finite direct products in which each factor is characteristic.

Lemma: Let be groups such that (c.f. §5.1 #2) is characteristic in for each . Then . Proof: We proceed by induction on . The conclusion is trivial for , and we proved the case above. Suppose the conclusion holds for some . Now . Suppose is an automorphism of ; then , so that is characteristic. By the induction hypothesis and part (a) of this problem, we have .

Now let be a finite abelian group. Note that every Sylow subgroup is normal, hence the unique subgroup of order . Thus each Sylow subgroup is characteristic in . In §5.4 #10, we saw that a finite abelian group is (isomorphic to) the direct product of its Sylow subgroups; the final conclusion then follows from the lemma.

### Like this:

Like Loading...

*Related*

or leave a trackback:

Trackback URL.