Decomposition of the gauge group of a sigma model I'm following the Chapter 5 (specifically, section 5.4) of Quigg's book Gauge Theories of the Strong, Weak and Electromagnetic Interactions and am confused with the following: 
Studying a sigma model with meson fields, he states that it is useful to 'decompose' the gauge group $SU(2)_L \times SU(2)_R$, with elements
$$
G_L=\exp{\left( i\pmb{\alpha}_L \cdot \frac{\pmb{\tau}}{2}\right) }; \;\;\;\;\ G_R=\exp{\left( i\pmb{\alpha}_R \cdot \frac{\pmb{\tau}}{2}\right) }
$$
'in terms of  $SU(2)_V \times SU(2)_A$', with elements
$$
G_I=\exp{\left( i\pmb{\alpha} \cdot \frac{\pmb{\tau}}{2}\right) }; \;\;\;\;\ G_5=\exp{\left( i\gamma_5\pmb{\alpha}_5 \cdot \frac{\pmb{\tau}}{2}\right) }.
$$
where $\pmb{\alpha}=\pmb{\alpha_R}+\pmb{\alpha_L}$ and $\pmb{\alpha}_5=\pmb{\alpha}_R-\pmb{\alpha}_L$.
I want to show that, after one vacuum is favored, the $SU(2)_V$ symmetry breaks, to which (I think) I need the explicit form of said decomposition.
Since the arguments of the exponentials do not commute (because the Pauli-matrices are weighted differently by the gauge parameters) I am not seeing how to write anything with this goal in mind. 
How can I do that? Or, if it is not needed, how can I show what I want?
More specifically, he defines a 2x2 mesonic spinless matrix field which transforms as
$$
\Sigma \to \Sigma'=G_L \Sigma G_R^{\dagger}.
$$
He then proceeds to write the field in terms of 4 scalars in the sigma(tau)-matrices basis:
$$
\Sigma=\sigma+i\pmb{\pi}\cdot\pmb{\tau}.
$$
The claim is that the vacuum state
$$
\left<\sigma\right>_0=v, \;\;\;\;\;\;\; \left<\pi\right>_0=\left(\begin{matrix}
0 \\
0 \\
0 \end{matrix}\right)
$$
(here $v$ is the VEV of the interaction) is unchanged by isospin rotations but breaks $SU(2)_A$ symmetry. 
That is what I want to show.
 A: Quite simply, under an $SU(2)_V$ (isospin) rotation,  $\pmb{\alpha}_5=0$, so that  $\pmb{\alpha_L}=\pmb{\alpha}/2=\pmb{\alpha_R}$, and
$$
G_L=\exp{\left( i\pmb{\alpha}_L \cdot \frac{\pmb{\tau}}{2}\right) }=G_R.  
$$
Consequently,
$$
\langle\Sigma\rangle=v  1\!\!1 \to  v G_L  1\!\!1 G_L^{\dagger}= v  1\!\!1.
$$
The vacuum is invariant under isospin.
By contrast, under the three axial rotations (there is no such thing as an $SU(2)_A$, and its promulgators deserve censure and ridicule for confusing impressionable minds),    $\pmb{\alpha}= 0$ so that $\pmb{\alpha}_5 /2=-\pmb{\alpha}_L=\pmb{\alpha}_R$, so that $G_R=G_L^\dagger$,
$$
\langle\Sigma\rangle=v  1\!\!1 \to  v G_L  1\!\!1 G_L=  v G_5^\dagger \neq v  1\!\!1.
$$ 
That means that axial transformations have shifted the vacuum, so the vacuum is not unique and invariant.


*

*Note I desist from assigning a group to the $G_5$s, since they cannot close to a group--in sharp contrast to the $G_I$s. You multiply $\Sigma$ on the left and the right by the same transformation matrix. But $G_5(\alpha)G_5(\beta)\neq G_5(\beta)G_5(\alpha)$ for a succession of such operations.(If you track down the details, you actually shift the pions by a constant--the essence of the celebrated Nambu-Goldstone realization.) 

*There is no chimera like $SU(2)_A$, and damn your misleading text which should know better. (These three generators are in what is technically known as a cosets space, and they correspond one-to-one with the massless pions of that model, the goldstons. I can see M G-M, the coinventor of that model, screaming at the mathematical solecism... But, trust me, it is not just math....)
