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My book uses this definition:

Let $G$ be a group. A subgroup $H$ of $G$ is called a characteristic subgroup if $\phi(H) \subseteq H$ for all $\phi \in \operatorname{Aut}(G)$.

But after some googling around, it seems that the definition for a characteristic subgroup involves equality $\phi(H) = H$.

Does $\phi(H) \subseteq H$ $\Rightarrow$ $\phi(H) = H$ ???

I tried to multiply by $\phi ^{-1}$ to get $H \subseteq \phi^{-1}(H)$ but I'm not sure if I am allowed to that and I'm even more unsure if $\phi(H) \subseteq H$ and $H\subseteq \phi^{-1}(H)$ $\Rightarrow$ $\phi(H) = H$.

Any mathematical wisdom? Thank you.

user1729
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abe
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  • You can apply $\phi^{-1}$ to both sides of $\phi(H)\subseteq H$ and get $H\subseteq\phi^{-1}(H)$. However this does not imply $\phi(H)=H$. – anon May 05 '14 at 02:24

3 Answers3

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For finite groups $\phi(H)\subseteq H\Rightarrow \phi(H)=H$ because $\phi$ is a bijection so $|\phi(H)|=|H|$ and no finite set can properly contain a set of the same size. Not so with infinite sets though.

It is possible for $\phi(H)$ to be contained strictly in $H$, in fact $\phi$ can be an inner automorphism (so that $H$ strictly contains one of its conjugates). For instance, let $\psi$ act on $\Bbb Q$ by $x\mapsto 2x$ and form the semidrect product $\Bbb Q\rtimes\langle\psi\rangle$. Let $H$ be the subgroup generated by $1\in\Bbb Q$. Then $\psi H\psi^{-1}=2H$ is strictly contained in $H$. Google "subgroup strictly contains conjugate" for more results.

anon
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The definitions are equivalent.

The answer of blue answers the question you ask, but does not address the underlying question of whether the two definitions are equivalent. So I will do that here.

So, no, $\phi(H) \subseteq H$ does not imply $\phi(H) = H$. However:

If $\phi(H) \subseteq H$ for all $\phi\in\operatorname{Aut}(G)$ then $\phi(H)=H$ for all $\phi\in\operatorname{Aut}(G)$. (Note: The key point is the "for all".)

Clearly, this implies that the definitions are equivalent. To see that the above holds, note that the following both hold for all $\phi\in\operatorname{Aut}(G)$.

  • $\phi(H)\leq H$

  • $\phi^{-1}(H)\leq H$

However, the second point can be manipulated to give you that $H\leq\phi(H)$. Combining this with the first point implies that $\phi(H)=H$ for all $\phi\in\operatorname{Aut}(G)$, as required.

Community
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user1729
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Well, $\phi(H)=H\implies \phi(H)\subseteq H$. The converse is false in general.