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The title pretty much sums it up.

$\operatorname{Inn}(G)$ is the group of inner automorphisms, $Z(G)$ is the center.

I know that $\operatorname{Inn}(G)$ is isomorphic to $G/Z(G)$. This means that we have a central extension (exact sequence):

$$1\to Z(G) \hookrightarrow G \to \operatorname{Inn}(G) \to 1. $$

The map between $Z(G)$ and $G$ is inclusion. The map between $G$ and $\operatorname{Inn}(G)$ is conjugation.

Is $\operatorname{Inn}(G) \cong G/Z(G)$ enough for the sequence above to be split? And most important, why/why not?

Thank you!

(Oh, also: does anybody know a book on this? Wikipedia is quite unclear. Thanks again.)

geodude
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    No, in general, there is no reason why it should split unless you assume some extra things. – Tobias Kildetoft Jul 23 '13 at 18:28
  • In most cases, I need G to be a connected Lie group. I don't know if this helps. – geodude Jul 23 '13 at 18:34
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    Also, there's no natural operation of $\operatorname{Inn}(G)$ on $Z(G)$ (except the operation by conjugation, which is trivial by definition adn woul dmake the semidirect product even a direct product) – Hagen von Eitzen Jul 23 '13 at 18:36
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    split central = direct. No. A group $G$ is isomorphic to the semidirect product $Z(G) \rtimes G/Z(G)$ iff it is a direct product of an abelian and a centerless group. This does happen for GL over some fields in some ranks, but is definitely not a feature in general. SL is more or less never like this. – Jack Schmidt Jul 23 '13 at 18:36
  • @JackSchmidt, you should make your comment into an answer. I'd like to +1 it. – Andreas Caranti Jul 24 '13 at 13:38
  • @AndreasCaranti: done. I wanted to find a connected Lie group example, and forgot that SL(2,Z) is a weird group. SL(4,Z) was much easier. – Jack Schmidt Jul 24 '13 at 16:01

3 Answers3

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No, this is not true in general. For example in the quaternion group, we have $\operatorname{Inn}(Q_8) \cong C_2 \times C_2$ but $Q_8$ does not contain a copy of $C_2 \times C_2$. Thus $Q_8$ is not a semidirect product of $Z(Q_8)$ and $\operatorname{Inn}(Q_8)$.

Mikko Korhonen
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Following m.k.'s answer. Any non-abelian nilpotent group is a counterexample. Why? We know that $Z(G/Z(G))$ is non-trivial in this case, therefore there is a non-trivial element $x$ in the center of $Inn(G)$. Therefore $G$ cannot be a direct sum of $Z(G)$ and $Inn(G)$ (since $x$ is in both).

Marci
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No, this is not true in general, even for connected Lie groups.

The special linear group $\operatorname{SL}_4$ in dimension 4 over a ring of characteristic not 2 (e.g. $\mathbb{R}$, resulting in a connected Lie group) is not a direct product of $Z(G)$ and $G/Z(G)$. This is because $-I$ is a commutator and in the center, but in $A \times B$ with $A$ abelian, the commutator subgroup is $1 \times [B,B]$, so that $[G,G] \cap Z(G) = 1$ in any such group with $A=Z(G)$. Explicitly, $$\tiny \left(\begin{array}{rrrr} -1& 0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0&-1& 0 \\ 0& 0& 0& 1 \\ \end{array}\right)^{-1} \cdot \left(\begin{array}{rrrr}% 0&1&0&0\\% 1&0&0&0\\% 0&0&0&1\\% 0&0&1&0\\% \end{array}\right)^{-1} \cdot \left(\begin{array}{rrrr} -1& 0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0&-1& 0 \\ 0& 0& 0& 1 \\ \end{array}\right) \cdot \left(\begin{array}{rrrr}% 0&1&0&0\\% 1&0&0&0\\% 0&0&0&1\\% 0&0&1&0\\% \end{array}\right) = \left(\begin{array}{rrrr}% -1&0&0&0\\% 0&-1&0&0\\% 0&0&-1&0\\% 0&0&0&-1\\% \end{array}\right)% $$

In fact, we can give a fairly tautological classification of the groups $G$ with $G \cong Z(G) \rtimes G/Z(G)$. Since $Z(G)$ is central, the semi-direct product is actually direct, $G \cong Z(G) \times G/Z(G)$, but the center of $A \times B$ is $Z(A) \times Z(B)$, so we must have $Z(G/Z(G)) = 1$. Hence $G$ is of the form $A \times B$ where $A$ is abelian and $B$ is centerless. Conversely, if $G$ if of the form $A \times B$ where $A$ is abelian and $B$ is centerless, then $Z(G) = A \times 1 \cong A$ and $G/Z(G) \cong B$, as required.

Proposition: $G \cong Z(G) \times G/Z(G)$ iff $G \cong A \times B$ where $A$ is abelian and $B$ is centerless.

Jack Schmidt
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