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Let $\alpha$ be a functor from the category of groups in the category of groups which assigns to every group $G$ a characteristic subgroup $\alpha (G)$ of $G$ and to every homomorphism $\theta : H \rightarrow K$ the restriction $\theta|_{\alpha(H)}:\alpha(H)\rightarrow \alpha(K)$ (in other words, $\alpha(H)^\theta \leq \alpha(K)$). So we have, in particular, that $\alpha(H)^\theta \leq \alpha(H^\theta)$. Is it true that also happen that $\alpha(H)^\theta$ is equal to $\alpha(H^\theta)$? I think it holds, but why?


D.J.S. Robinson, "Finiteness Conditions and Generalized Soluble Groups", Vol. 1 Pag. 18, (1.3, at the bottom of the page)

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There are many characteristic subgroups (e.g. all subgroups of $G$ is $G$ is cyclic), so $\alpha$ might make different "choices" for different $G$. Do I understand the statement right that the only restriction to these choices is that $\alpha(H)^\theta \le \alpha(K)$ for each homomorphism $\theta\colon H\to K$? – Hagen von Eitzen Jan 14 '14 at 15:55
Yes, but I think that this isn't a real restriction since otherwise $\alpha$ could not be a functor assigning the restriction to each homomorphism. – W4cc0 Jan 14 '14 at 15:57
Yes but you need this assumption (which is stronger than being characteristic!) in order to get a functor. – Martin Brandenburg Jan 14 '14 at 16:33
1 and are the papers of Baer mentioned. Baer appears to have switched from functor in the first to Robinson's definition in the second. Presumably Baer also found the equivalence obvious. – Jack Schmidt Jan 14 '14 at 16:42
@JackSchmidt this is why I asked my question. It seems that both Robinson and Baer found the equivalence obvious. But to me is not so obvious :(. I "read" the paper of Baer but I cannot realize anything new. – W4cc0 Jan 14 '14 at 16:51
up vote 4 down vote accepted

It is not true at least for the category of Abelian groups. Take $\alpha (G)$ the subgroup of all elements of finite order. Set $H=\mathbb{Z}$ and $K=\mathbb{Z}_n$.

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But in this case: $\alpha(H)^i=[H,H]^i=\{1\}^i=\{1\}=[H^i,H^i]$ since $H$ and $H^i$ are both abelian (where $i$ is the identity function of $H$ in $K$). So this isn't a counterexample I think. – W4cc0 Jan 14 '14 at 16:08
Sorry, I read the question wrong. I changed the answer. – Boris Novikov Jan 14 '14 at 16:39
The subgroup generated by the elements of finite order might work in general. – Jack Schmidt Jan 14 '14 at 16:54
In your counterexample (following my notation) I think must be $K$ changed by $H$. Anyway it seems to work also in general. So the statment in Robinson's book is false? – W4cc0 Jan 14 '14 at 17:08
I don't know, I have not this book. – Boris Novikov Jan 14 '14 at 17:17

If you look at $\alpha(G)=G'$, the commutator subgroup, then it is true. In general verbal subgroup will do.

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Yes but only the verbal subgroup will do this? This means that if $\alpha$ satisfies to my requirements then $\alpha$ is a "verbal" mapping? – W4cc0 Jan 14 '14 at 16:31
Well, that is not clear. With respect to any variety of groups in any case you will have a lot of characteristic subgroups that are eligible, see for example – Nicky Hekster Jan 14 '14 at 16:34
Well, I know about verbal subgroup; my problem is the fact that Robinson in the text I cited says that: a group theoretical function (a function $\alpha$ assigning to each group $G$ a subgroup $\alpha(G)$ s.t. if $\theta : G \rightarrow K$ is an isomorphism of $G$ then $\alpha(G)^\theta=\alpha(G^\theta)$ is a functor of the category of groups if and only if for every homomorphism $\theta: G \rightarrow H$ it follows that $\alpha(G)^\theta=\alpha(G^\theta)$ - here the image of a homormorphism $G\rightarrow H$ under $\alpha$ is the restriction to $\alpha(G)$) – W4cc0 Jan 14 '14 at 16:41
I proved that if the condition holds then $\alpha$ is a functor but I cannot see why if $\alpha$ is a functor the condition hold. If the condition hold then it can be proved that $\alpha$ is a "verbal" mapping. @JackSchmidt Thanks, I removed it – W4cc0 Jan 14 '14 at 16:43

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