Degree of transitive constituents is odd implies $|G|$ is odd I want to prove that: the order of a permutation group $G \le S^\Omega$ is odd if and only if the degrees of all transitive constituents of $G$ and the degrees of all transitive constituents of each $G_\alpha (\alpha \in \Omega)$ are odd.  This is Exercise 3.13 in Wielandt's text.  
(Let me recall that if $\Delta \subseteq \Omega$ is a fixed block of $G$, then $G$ restricted to $\Delta$, denoted $G^\Delta$, is called a constituent.  It is transitive iff $\Delta$ is a minimal fixed block (i.e. an orbit).  So the assertion asks to show that $|G|$ is odd iff each orbit of $G$ and each orbit of each $G_\alpha$ has odd length.) 
The hint given is to use these three facts: (i) $|G_\alpha|~|\alpha^G|=|G|$. (ii) $|G:G_{\alpha \beta}| = |\alpha^G| |\beta^{G_\alpha}| = |\beta^G| |\alpha^{G_\beta}|$. (iii) If prime $p$ divides $|G|$, then $G$ contains an element whose cycle decomposition contains a $p$-cycle.
The implication is proved using (ii): If $|\alpha^G|$ or $|\beta^{G_\alpha}|$ is even for any $\alpha,\beta$, then by (ii) $|G|$ is even.  How do I prove the converse?
 A: The question is essentially answered in my comment and Derek Holt's comment above.  This answer elaborates on these two comments.
Let $G \le S^\Omega$. We need to show that if $|G|$ is even, then some orbit $\alpha^G$ of $G$ or some orbit $\beta^{G_\alpha}$ of some $G_\alpha$ $(\alpha \in \Omega)$ has even length. 
Since $|G|$ is even, by (iii) above there exist $g,\alpha,\beta$ such that $g=(\alpha,\beta) \cdots \in G$. It suffices to show that for this $\alpha,\beta$, $|G:G_{\alpha \beta}|$ is even; by (ii) it would then follow that one of $|\alpha^G|$ or $\beta^{G_{\alpha}}|$ is also even.  The permutation $g$ contains the 2-cycle $(\alpha,\beta)$.  Hence $g$ fixes the set $\{\alpha,\beta\}$ setwise but not pointwise, i.e. $g \in G_{\{\alpha,\beta\}} - G_{\alpha \beta}$.  Also, every element in $G_{\{\alpha,\beta\}}$ is either in the subgroup $G_{\alpha \beta}$ or in the coset $g G_{\alpha \beta}$, whence the index $|G_{\{\alpha,\beta\}}:G_{\alpha \beta}|$ equals 2. Thus, the index $|G:G_{\alpha \beta}| = |G:G_{\{\alpha,\beta\}}| |G_{\{\alpha,\beta\}}:G_{\alpha \beta}|$ is even.
