# On groups whose center has odd order

Let $G$ be a finite group such that $Z(G)$ is of odd order and $Inn(G)$ is of even order. Then prove $G\simeq Z(G)\times N$, such that $N$ is a subgroup of $G$ where $N\simeq Inn(G)$. Thank you!

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It is not possible to prove false statements. –  Derek Holt Dec 30 '12 at 18:23
@maryam, where are these questions from? Some book...? –  DonAntonio Dec 30 '12 at 18:32

Let $V$ be the Klein $V$ group and $A$ be the cyclic group of order $9$, generated by $a$. Let $\rho:A\rightarrow \text{Aut}(V)$ be defined by $$\rho:a\mapsto \left(\begin{array}{cc}1&1\\1&0\end{array}\right).$$ Now we take $G=V\rtimes_\rho A$.
Then the center of $G$ is $\text{ker}(\rho)$, which has order $3$. But $Z(G) \times \text{Inn}(G)\cong \ker{\rho}\times G/\text{ker}(\rho)$ has no element of order $9$. So your claim is false.
(Edit: Note that we know such a $\rho$ exists simply because $3$ divides $|GL_2(\mathbb{F}_2)|=(2^2-1)(2^2-2)$. But I have given $\rho$ explicitly for the sake of completeness.)