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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?

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  • $\begingroup$ 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
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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.

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