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This is a sequel to Why can the homomorphism $\phi$ in semi-direct products only be varied by inner automorphisms upon changing the complement group?

My professor made another remark that:

Let's go back to the extension question

$1 \to A \to G \to B \to 1$

It turns out that a general extension of $B$ by $A$ is a mutual generalization of two extremes. One extreme is a semidirect product, in which $A$ is complemented but conjugation by $B$ (and therefore by elements of $G$ in general) can be highly non-trivial. In the other extreme, $A$ is a subgroup of the center $Z(G)$, so that conjugation in $G$ does nothing whatsoever to $A$; but the search for a complement of $A$ may be an utter failure. This extreme is called a central extension, and it is another construction that you should learn in the context of both Jordan-Holder and p-groups.

Questions:

  1. In case of central extensions, why exactly can the search for a complement of $A$ be an utter failure?

  2. What is meant when he says that conjugation by $B$ can be highly non-trivial in case of semi-direct products?

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  • $\begingroup$ It's a strange expression to use, but it can be an utter failure because there isn't one to find! If you dig a hole in the group searching for gold that can be an utter failure too. $\endgroup$
    – Derek Holt
    Apr 14, 2020 at 7:46
  • $\begingroup$ @DerekHolt So why can't a complement of $A$ be found, in the case of central extensions? $\endgroup$
    – user568976
    Apr 14, 2020 at 7:56
  • $\begingroup$ As I just said, it cannot be found because there is no complement to find. I don't really understand what you are asking. $\endgroup$
    – Derek Holt
    Apr 14, 2020 at 9:01
  • $\begingroup$ @DerekHolt I'm asking "why is there no complement to find?". Is it a characteristic property of central extensions (that is, complements of subgroups of the center do not exist)? "It cannot be found because there is no complement to find" is a tautology that doesn't really answer this. $\endgroup$
    – user568976
    Apr 14, 2020 at 9:09
  • $\begingroup$ I am afraid I continue to think that your question is equivalent to asking why you cannot find gold if you dig a hole in your garden, but I have written an answer with examples. $\endgroup$
    – Derek Holt
    Apr 14, 2020 at 10:36

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I believe that this definition of "central extension", as one with $A \le Z(G)$ in which there is no complementis non-standard. The standard meaning of "central extension" is simply one in which $A \le Z(G)$, irrespective of whether there are complements are not.

But let me give you three examples in which $A \le Z(G)$. In the first of these there is a complement, and in the second and third there is not.

Example 1. Let $G = \langle x \rangle$ be cyclic of order 6: so $x^6=1$ (the identity). Let $A = \langle x^3 \rangle$ be the subgroup of $G$ of order $2$, and $B \cong G/A$ is cyclic of order $3$. So $A \le Z(G)$: in fact $G = Z(G)$. Now let $C$ be the subgroup $\langle x^2 \rangle$ of $G$, which is cyclic of order 3 with $C \cong G/A$. Then $B$ is a complement to $A$ in $G$: in fact it is the unique such complement.

Example 2. Now let $G = \langle x \rangle$ be cyclic of order 4: so $x^4=1$. Let $A = \langle x^2 \rangle$ be the subgroup of $G$ of order $2$, and $B \cong G/A$ is cyclic of order $2$. Again we have $A \le Z(G)$ and $G = Z(G)$. This time there is no complement $C$ of $A$ in $G$. Such a complement would have to be isomorphic to $B$ (i.e. cyclic of order 2) and satisfy $A \cap C = \{1\}$. But $A$ is the only subgroup of $G$ of order $2$, so no such $C$ exists.

Example 3. Let $G = \langle 1,r,r^2,r^3,s,sr,sr^2,sr^3 \}$ be the dihedral group order $8$. That is the group of rotations and reflections of a square, where $r,r^2,r^3$ are rotations, and $s,sr,sr^2,sr^3$ are reflections. Let $A=Z(G)$, which is the subgroup $\langle r^2 \rangle$ of order $2$. Note that $B \cong G/A$ is a Klein four group. A complement $C$ of $A$ in $G$ would be a subgroup of order $4$ isomorphic to $B$ with $A \cap C = \{1\}$. There is no such subgroup. There are subgroups such as $\{1,r^2,s,r^2\}$ (in fact there are two such subgroups) that are isomorphic to $B$, but they all contain $r^2$.

As an exercise, try to think of an example in which $A = Z(G)$ and there are complements of $A$ in $G$.

I am afraid that I do not really know the answer to your second question. It seems to me like an informal comment, and I would advise you not to worry about it.

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  • $\begingroup$ I don't think the question suggests that "central extension" is reserved for the cases where there is no complement. Just that it might happen. $\endgroup$ Apr 14, 2020 at 10:42
  • $\begingroup$ @CaptainLama The quote says "This extreme is called a central extension" which appears to be referring to the situation where the search for a complement proved to be an utter failure. I don't find this informal language very helpful. Perhaps my search for a complement was an utter failure because I am very poor at finding complements? $\endgroup$
    – Derek Holt
    Apr 14, 2020 at 11:15

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