2
$\begingroup$

Suppose $G_1$ is a group, which has no outer automorphism.

Suppose $G_2$ is a group, which has no outer automorphism.

Main Question: What are the possible conditions to know can there be an outer automorphism for $$ G_1 \times G_2? $$

An easier question:

Suppose $G_1=SU(2)$ is a group, which has no outer automorphism.

Suppose $G_2= \mathbb{Z}/2\mathbb{Z}$ is a group, which has no outer automorphism.

  • $SU(2) \times SU(2)$ has however a order-2 outer automorphism.
  • $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ has however a order-2 outer automorphism.

Simpler Question: Is there an outer automorphism for $$ SU(2) \times \mathbb{Z}/2\mathbb{Z} ? $$

If you can answer anything of the above two Questions - this counts as a perfect answer.

$\endgroup$
3
$\begingroup$

A sufficient condition is that $G_1$ has a central subgroup $C$ isomorphic to a quotient of $G_2$ - this holds for your example.

We then have a natural map $q:G_2\to C$.

The map $\phi:G_1\times G_2\to G_1\times G_2:(g,h)\mapsto (gq(h),h)$ is an outer automorphism.

$\endgroup$
  • $\begingroup$ How do we show this? thanks +1!! $\endgroup$ – annie heart Feb 25 at 15:53
  • $\begingroup$ just check the definition of automorphism. Injective: $\phi(g,h)=(1,1)$ then immediately $h=1$ so $q(1)=1$ gives $g=1$. Surjective: $(g,h)=\phi(gq(h)^{-1},h)$. Homomorphism: A little long for a comment, but straightforward $\endgroup$ – Robert Chamberlain Feb 25 at 16:04
  • 1
    $\begingroup$ Should it be intead a natural map $q:G_2\to C$. The map $\phi:G_1\times G_2\to G_1\times G_2:(g,h)\mapsto (gq(h),h)$ is an outer automorphism??? $\endgroup$ – wonderich Feb 26 at 15:55
  • $\begingroup$ yep, typo, well spotted $\endgroup$ – Robert Chamberlain Feb 26 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.