Let $A$ be a finite dimensional central simple algebra over a field $F$ of characteristic $0$. So by Weddernburn's theorem, $A\cong M\_n(D)$ for some division algebra $D$ over $F$. Let $\dim\_F(D)=m^2$. Then $m$ is called the index of $A$. 


Assume that $A$ is crossed product: there is a finite Galois extension $E$ of $F$, say of degree $n$, which is (isomorphically sitting) inside $A$ and $n$ invertible elements $u\_1,\ldots, u\_n\in A$ such that $Eu\_1\oplus \cdots \oplus Eu\_n=A$ (direct sum as $F$-vector space). The product of elements of $A$ from this decomposition are determined by an equivalence class of $2$-cocycle $f$ in $H^2({\rm Gal}(E/F), E^\*)$ (together with action of Galois group on $E^\*$). 


**Q.** In general, it is known that the order of $[f]$ in $H^2({\rm Gal}(E/F), E^\*)$ divides the index $m$ of $A$. Further, if $F$ is a number field, then equality holds. So, what is an example where order of $[f]$ properly divides the index $m$ of $A$?


The results mentioned in question are taken from Albert's book Structure of Algebras.



Example of a central simple algebra
You can take: $F=\mathbb{C}(X\_1,Y\_1,\ldots,X\_n,Y\_n)$ and take the tensor product of quaternion algebras $$A=(X\_1,Y\_1)\_F\otimes\_F\cdots\otimes (X\_n,Y\_n)\_F.$$ Here $A$ contains a subfield $E$ isomorphic to $F(\sqrt{Y\_1},\ldots \sqrt{Y\_n})$ for example.


Now $A$ is a division algebra, so it has index $2^n$, but it has order $2$ in $\mathrm{Br}(E/F)\simeq H^2(\mathrm{Gal}(E/F),E^\times)$.


PS. If you need details, I can post something this evening.


**Edit.** Here is a proof that $A$ is division. I will assume some knowledge on quadratic forms. Note that I replace $\mathbb{Q}$ by $\mathbb{C}$ to get rid of some ennoying coefficients (see below), but the result is the same.


If $A$ is a central simple algebra over a field $F$, let $T\_A:A\to F$ the quadratic form defined by $T\_A(a)=Trd\_A(a^2)$. 


**Lemma.** If $T\_A$ is anisotropic, then $A$ is division.


*Proof.* Assume $A$ is not division, so $A\simeq M\_r(D)$, where $D$ is division. The matrix $a\_0$ filled with zero, except on the top right corner where the entry is $1\_D$, satisfies $a\_0^2=0$, hence $T\_A(a\_0)=0$, while $a\_0\neq 0$. Hence, $T\_A$ is isotropic.


**Prop.** Let $F$ be a field of characteristic different from $2$. Assume that $q\_1,q\_2$ are anistropic quadratic forms over $F$. Then the form $q\_1+Xq\_2$ is anistropic over $F(X)$.


*Proof.* Omitted. This is a standard result.


The rest of the proof is devoted to show that the trace form of our example is anisotropic.


**Easy fact.** We have $T\_{A\otimes B}\simeq T\_A\otimes T\_B$ (by standard properties of the reduced trace)


Now if $Q=(a,b)\_F$, the standard basis $1,i,j,ij$ is orthogonal wrt $T\_Q$, and $$T\_Q\simeq \langle 2,2a,2b,-2ab\rangle\simeq \langle 1,a,b,ab\rangle\simeq \langle 1,a\rangle\otimes \langle 1,b\rangle.$$ 
The second isomorphism comes from the fact that $2$ and $-2$ are squares in $\mathbb{C}$; this is why I changed.


Hence $T\_A\simeq \langle 1,X\_1\rangle\otimes \langle 1,Y\_1\rangle\otimes\cdots\otimes \langle 1,X\_n\rangle\otimes \langle 1,Y\_n\rangle$.


Using the previous proposition and induction (starting form the fact that $\langle 1\rangle$ is anisotropic), we get the desired result.


Of course, you can do all of this for $n=2$...


