Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
2  
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
2  
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
7  
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
show 1 more comment

3 Answers

up vote 5 down vote accepted

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.

share|improve this answer
add comment

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 $\operatorname{Inn}(Q_8)$.

share|improve this answer
add comment

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.