Let $\{P\_i\}$ be a subset of $SU(n)$ such that for any $U$ in another subset (or perhaps subgroup) $H$ of $SU(n)$: $$P\_1UP\_2U\cdots P\_mU=I$$ where $I$ is the identity element. Is there a sequence $\{P\_i\}$ such that $H$ be enlarged to the whole $SU(n)$? Also (perhaps more interesting): how large can $H$ be made?


Perhaps a similar question can be asked about groups in general. For $SU(n)$, I suspect that $U$'s must be generated by a single Lie Algebra element [which I understand to mean $U=exp(-i \theta A)$ for a fixed Hermitian $A$ and any real number $\theta$]. Notice that a similar question can be asked about rotation matrices in $\mathbb{R}^3$ which I believe to be sufficient for a more general proof. A direction that I pursued was to note that the transformation of the generator (logarithm) of $U$ under the sequence has to be a linear map (identically zero) and then try to prove that it (or its restriction) can have a dimension of at most 1.


As a non-$SU(n)$ example consider the group of affine transformations of the 3-dimensional Euclidean space and take ${P\_i}$ to be 4 alternating 180 rotations (flips) along $x$ and $y$. Then $U$ can be any translation. Now It is highly unlikely that the set of $U$'s can be extended to the whole affine group by using a larger set of $P\_i$'s but I can't prove it. 


Excuse my quantum physicist notation and I hope that this is not too elementary although I would appreciate an elementary answer. 



Multiplicative Identity for all elements in SU(n)
I can show that it is impossible to achieve $H=SU(n)$. More generally, if $G$ is a compact connected Lie group, I will show that the map $U \mapsto P\_1 U P\_2 U \cdots P\_m U$ is surjective, and therefore the preimage of the identity can't be all of $G$.


**Proof** Such a $G$ is a compact connected orientable manifold so we can talk about the degree of self-maps of $G$. If $\phi: G \to G$ is a continuous function which misses a point, then $\phi$ has degree $0$. We will compute the degree of $U \mapsto P\_1 U P\_2 U \cdots P\_m U$ and see that it is nonzero.


Since $G$ is connected, $U \mapsto P\_1 U P\_2 U \cdots P\_m U$ is homotopic to $U \mapsto U^m$. A generic matrix in $SU(n)$ has $m^{n-1}$ many $m$-th roots in $SU(n)$ (and $m^n$ many $m$-th roots in $U(n)$). At some point in my life, I worked out that all of these $m$-th roots contribute to the degree with the same sign. More generally, let $G$ be a compact connected Lie group of rank $r$ (meaning that maximal torii have degree $r$). For a generic element $x$ of $G$, there is a unique maximal torus $T$ containing $x$ and all $m$-th roots of $x$ are in that torus; there are $m^r$ of them. For our purpose, the only important fact is that $m^r \neq 0$, so $U \mapsto U^m$ has nonzero degree and $U \mapsto P\_1 U P\_2 U \cdots P\_m U$ must be surjective. $\square$


**ADDED** Robert Bryant proves that, if $G$ is a compact connected Lie subgroup and $T$ is a connected subgroup so that $P\_1 U P\_2 U \cdots P\_m U=1$ for $U \in T$, then $T$ is contained in a maximal torus. The point of this edit is to point out that, if $G$ is semisimple, and $T$ is any maximal torus, then there are $P\_i$ which achieve this. Robert Bryant already does this for $SU(n)$ in his answer. 


Let $N(T)$ be the normalizer of $T$, so we have a short exact sequence $1 \to T \to N(T) \to W \to 1$, where $W$ is the Weyl group. Let $c$ be any element of the Weyl group which acts with no eigenvalues of $1$ on the Lie algebra of $T$ (for example, a Coxeter element) and let $h$ be the order of $c$. Let $P\_1$ be a lift of $c$ to $N(T)$. Let $P\_1^h = z$ (an element of $T$). Set $P\_1=P\_2 = \cdots = P\_{h-1}$ and $P\_h = P\_1 z^{-1}$. Then I compute
$\def\Ad{\mathrm{Ad}}$
$$P\_1 \exp(t) P\_1 \exp(t) \cdots P\_1 \exp(t) P\_1 z^{-1} \exp(t) = \exp\left( (\Ad(P\_1) + \Ad(P\_1)^1 + \cdots + \Ad(P\_1)^h ) t \right) P\_1^h z^{-1} = \exp(0) \cdot 1.$$


