We say that a space $X$ *has a homology $p$-exponent* if some power of $p$ annihilates the $p$-torsion in $H\_\ast(X;\mathbb{Z})$.


I am interested in the homology exponents of the free infinite loop space $QX=\Omega^\infty\Sigma^\infty X$. Apparently, by examining the Bockstein spectral sequence, it should be possible to determine the integral homology of $QX$ as a functor of the integral homology of $X$ (see May's [Homology operations on infinite loop spaces](http://www.math.uchicago.edu/~may/PAPERS/11.pdf), Remark 2.6). Before I get into the gory details of this calculation, however, I would like to ask if the following is known.



> 
> If $X$ is a space with a homology $2$-exponent, must $QX$ have a homology $2$-exponent also?
> 
> 
> 



Homology exponents for $QX$
By May's Remark 2.6, the answer should be no. If $X$ is a Moore space with
$\tilde H\_\*(X; Z) = Z/p$ concentrated in an odd degree $2q-1$, and
$\tilde H\_\*(X; Z/p) = Z/p\{x, y\}$ with $\beta\_1(y) = x$, then in
$H\_\*(Q(X); Z/p)$ the $r$-th Bockstein $\beta\_r$ is
defined and nonzero on $y^{p^{r-1}}$ for each $r\ge1$, so $H\_\*(Q(X); Z)$ contains a copy of $Z/p^r$ in degree $2p^{r-1}q-1$. Hence there is no uniform power of
$p$ that annihilates all the $p$-torsion.


