A question about complements in a group I consider the property for a group $G$, that every time you take an element $g$ of prime order in $G$, then there is a complement $H$ to $\langle g\rangle$ in $G$ in the sense that $G=\langle g\rangle H$ and $\langle g\rangle\cap H=1$.
Let $G$ be the cartesian product (= unrestricted direct product) of infinitely many (say countable for simplicity) copies of $S_3$. It is very easy to show that $G$ satisfies this property.
I've read on some Russian paper that actually this property holds in all factor groups $G/N$ of $G$, but I'm unable to prove this.
I can reduce easily to the case in which $N\leq G'$ and $\langle gN\rangle\leq G'/N$ but I do not know how to go on from this. Any suggestion?
 A: It's true (I initially claimed the contrary): the case of elements of order $2$ is easy and the case of elements of order $3$ is a little more tricky.
Write $G=S_3^I$. Let $G_3=C_3^I$ be the 3-Sylow subgroup in $G$. Fix a 2-Sylow subgroup $C_2$ in $G_3$ and write $G_2=C_2^I$. We consider a quotient $Q=G/N$ of $G$.
Write $Q_3$ and $Q_2$ for the image of $G_3$ and $G_2$ in $Q$. Then $Q_p$ is elementary $p$-abelian, $Q=Q_2Q_3$ with $Q_3$ normal, and $Q_2\cap Q_3=\{1\}$. Hence $Q=Q_3\rtimes Q_2$.
(a) Suppose $c\in Q$ has order $2$; let $d$ be a lift of $c$ in $G$; then $d$ has order $2$ or $6$ and so up to replace $d$ with $d^3$, we can suppose that $d^2=1$; then up to conjugate in $G$, we can suppose that $d\in G_2$ and hence that $c\in Q_2$. Then it's clear that $\langle c\rangle$ has a normal complement (first mod out by $Q_3$ and boil down to the case of an elementary abelian 2-group).
(b) Let $\mathcal{T}$ be an ideal of subsets of $I$ (i.e., a set of subsets of $I$ with $\emptyset\in \mathcal{T}$, $\mathcal{T}$ stable under taking finite unions, and intersections with subsets of $I$). Define $G_3(\mathcal{T})=\bigcup_{J\in \mathcal{T}}C_3^J$: this is a subgroup of $G_3=C_3^I$; moreover it is a $(\mathbf{Z}/2\mathbf{Z})^I$-submodule, i.e., it is stable under its obvious action switching the sign of coordinates, so that the corresponding semidirect product is $G$. If $\mathcal{T}$ is the complement of an ultrafilter $\mathcal{U}$, then $G_3(\mathcal{T})$ is the kernel of the homomorphism $G_3=C_3^I\to C_3$ taking $(x_i)_{i\in I}$ to its $\mathcal{U}$-limit. 
Conversely, it is easy to check that every $(\mathbf{Z}/2\mathbf{Z})^I$-submodule of $G_3$ has the form $G_3(\mathcal{T})$ for some ideal $\mathcal{T}$ of subsets of $I$. Indeed, the $(\mathbf{Z}/2\mathbf{Z})^I$-submodules of $G_3$ of index 3 are precisely the $G_3(\mathcal{T})$ for $\mathcal{T}$ the complement of some ultrafilter. 
Now let $c\in Q$ have order $3$. Let $d\in G$ be a lift of $c$: then $d$ has order $3$ or $6$, and replacing $d$ with $d^4$ if necessary we can suppose that $d$ has order $3$. The support of $d$ is a nonempty subset $J$ of $I$. By the above, $N\cap G_3=G_3(\mathcal{T})$ for some ideal $\mathcal{T}$ of subsets of $I$ (so $J\notin \mathcal{T}$). Let $\mathcal{M}$ be a maximal ideal (of the Boolean algebra of subsets of $I$) such that $\mathcal{T}\subset \mathcal{M}$ and $J\notin \mathcal{M}$. (So the complement $\mathcal{U}$ of $\mathcal{M}$ is an ultrafilter.)
Then $d\notin G_3(\mathcal{M})N$. Indeed, we would have otherwise $d=gn$ in this decomposition, so $n=dg^{-1}\in N\cap G_3\subset G_3(\mathcal{M})$ and in turn $d\in G_3(\mathcal{M})$, a contradiction. Hence it is no restriction to assume that $N\cap G_3$ has index $3$ in $G_3$. Then the result follows from the following:
Claim: let $Q$ be a group with a normal subgroup $Q_3$ of order $3$ such that the quotient $Q/Q_3$ is elementary 2-abelian. Then $Q_3$ has a complement in $Q$.
Proof of the claim: let $A$ be the centralizer of $Q_3$. Then $A$ is characteristic in $Q$, and is the product of $Q_3$ with an elementary $2$-abelian subgroup $A_2$. Hence modding out by $A_2$, we can suppose that $Q_3$ equals its own centralizer. Then in this case $Q/Q_3$ acts faithfully on $Q_3$, so has order $1$ or $2$, so either $Q=Q_3$ and we're done, or $Q\simeq S_3$ and we're done too. 
