Let $G=H\times K$ and $H\times 1$ be a characteristic subgroup of $G$.

Then can we conclude that $1\times K$ is also a characteristic subgroup of $G$?

My motivation is the case where orders of $H$ and $K$ are relatively prime. In that case, both must be characteristic subgroups of $G$. So I wonder: if one of the components is characteristic in $G$ then is the other, too?

  • $\begingroup$ strictly speaking that $=$ should be $\cong$ $\endgroup$ – janmarqz Dec 29 '14 at 19:26
  • $\begingroup$ @janmarqz: I mean exactly the group $H\times K$ by saying $G$, why should I use $\cong$ ? $\endgroup$ – mesel Dec 29 '14 at 19:29
  • $\begingroup$ how is $H<H\times K$? $\endgroup$ – janmarqz Dec 29 '14 at 19:31
  • $\begingroup$ @janmarqz: ok, I edited. $\endgroup$ – mesel Dec 29 '14 at 19:33

The answer is no.

For example in $C_2 \times S_n$ ($n \geq 3$) the $C_2$ factor is characteristic because it is the center, but the $S_n$ factor is not characteristic: consider the automorphism $(x, \sigma) \mapsto (x \operatorname{sgn}(\sigma), \sigma)$.

With GAP you can find plenty of more examples.


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.