Multiplicative Identity for all elements in SU(n) 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. 
 A: 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
$\def\Ad{\mathrm{Ad}}$
$$P_1 \exp(t) P_1 \exp(t) \cdots P_1 \exp(t) P_1 z^{-1} \exp(t) = \exp\left( (\Ad(P_1) + \Ad(P_1)^1 + \cdots + \Ad(P_1)^h ) t \right) P_1^h z^{-1} = \exp(0) \cdot 1.$$
A: New answer:  I now have an answer for the subgroup case that the OP originally asked about.  In fact, one has the following result:  Let $G$ be a connected compact Lie group and let $p = (p_1,\ldots,p_m)$ be an $m$-tuple of elements of $G$.  Define the set
$$
H(p) = \{ g\in G\ |\ (p_1g)(p_2g)\cdots(p_mg) = 1\}.
$$
If $H\subset H(p)$ is a connected subgroup of $G$, then $H$ is abelian.  In particular, $H$ lies in a maximal torus in $G$.
I am going to give the proof in Part III below.  I am leaving Parts I and II (which were part of my previous answer) in place because they contribute to the answer below in some way.
Part I: In the case $G=\mathrm{SU}(n)$, one can have an $H\subset H(p)$ be a maximal torus for appropriate choice of $p$, so the abelian condition is the best possible.
Just do this:  Let $T^{n-1}\subset \mathrm{SU}(n)$ be the maximal torus that consists of diagonal elements, i.e., the elements in $\mathrm{SU}(n)$ that preserve the lines $L_i=\mathbb{C}e_i\subset\mathbb{C}^n$, where $e_i$ is the standard basis of $\mathbb{C}^n$. Let $Q\in\mathrm{SU}(n)$ be the matrix that satisfies $Qe_i=e_{i+1}$ for $1\le i<n$ and $Qe_n=(-1)^{n-1}e_1$.  (Note that $Q^n=(-1)^{n-1}I$.) Then, for $U\in T^{n-1}$, one has the identity
$$
U\cdot {Q}UQ^{-1}\cdot {Q}^{2}UQ^{-2}\cdot\ \cdots\ \cdot {Q}^{(n-1)}UQ^{-(n-1)} = I.
$$
(Each of the conjugates in the product is a diagonal matrix with the eigenvalues of $U$ cyclically permuted, and the product of the eigenvalues of $U$ is $1$.) Conjugating this identity by $Q^{n-1}$, one obtains
$$
Q^{1-n}U(QU)^{n-1} = I
$$
for all $U\in T^{n-1}$, which is an identity of the form the OP desired (with $m=n$).
Part II:  Here is an algebraic proof that one can't have $H=\mathrm{SU}(n)=G$:  Suppose that one had
$$
P_1UP_2U\cdots P_mU = I
$$
for all $U\in \mathrm{SU}(n)$.  Then, complexifying this relation, one would have
$$
P_1XP_2X\cdots P_mX = I
$$
for all $X\in \mathrm{SL}(n,\mathbb{C})$.  [Put another way: The mapping $f:\mathrm{SL}(n,\mathbb{C})\to \mathrm{SL}(n,\mathbb{C})$ defined by $f(X)=P_1XP_2X\cdots P_mX$ is holomorphic, but, by hypothesis, it is constant on the totally real submanifold $\mathrm{SU}(n)\subset \mathrm{SL}(n,\mathbb{C})$, which is the fixed point set of the antiholomorphic involution $X\mapsto {}^t\bar X^{-1}$. Thus, $f$ is constant.]
Multiplying by $\lambda^m\in\mathbb{C}^\ast$, one finds
$$
P_1(\lambda X)P_2(\lambda X)\cdots P_m(\lambda X) = \lambda^m\ I
$$
for all $X\in \mathrm{SL}(n,\mathbb{C})$ and $\lambda\in\mathbb{C}^\ast$, which implies that
$$
P_1ZP_2Z\cdots P_mZ = \det(Z)^{m/n}\ I
$$
for all $n$-by-$n$ complex matrices $Z$ with nonzero determinant. Since the left hand side is a polynomial in the entries of $Z$, it follows that $m/n = k$ for some integer $k$, so
$$
P_1ZP_2Z\cdots P_mZ = \det(Z)^k\ I
$$
for all $n$-by-$n$ complex matrices $Z$, where $m=kn$.  Now, choose $Z$ to be rank $1$, say, $Z=x\ {}^ty$ for $x,y\in \mathbb{C}^n$.  Taking the trace of the above relation and using the fact that $\det(Z)=0$, one finds that the following product of quadratic forms must vanish identically
$$
({}^tyP_1x)\ ({}^tyP_2x)\ ({}^tyP_3x)\ \cdots ({}^tyP_mx) = 0.
$$
Since $x$ and $y$ are arbitrary, it follows that one of the factors ${}^tyP_jx$ must vanish identically.  But this is clearly impossible, since each $P_j$ is invertible.
Remark: In fact, this proof works for any of the real forms of $\mathrm{SL}(n,\mathbb{C})$, such as $\mathrm{SL}(n,\mathbb{R})$, $\mathrm{SU}(k,n{-}k)$, or $\mathrm{SU}^\ast(n/2) = \mathrm{SL}(n/2,\mathbb{H})$, since they all have the same complexification.
Part III: Now, finally, the general proof that $H$ must be abelian.
First, by embedding $G$ into some $\mathrm{SU}(n)$ for some $n$ sufficiently large, I can assume that $G=\mathrm{SU}(n)$.  Next, if $p=(p_1,\ldots,p_m)$ were such that $H(p)$ contained a connected nonabelian subgroup $H\subset G$, then, by replacing $H$ by its closure, I can assume that $H$ is a connected compact nonabelian Lie subgroup of $G$.  In particular, it follows from the classification of compact Lie groups that $H$ contains a Lie subgroup $K$ that is isomorphic to either $\mathrm{SO}(3)$ or to $\mathrm{SU}(2)$.
Now recall that every compact simple Lie group $G$ has a closed, left-invariant Cartan $3$-form $\gamma_G$, whose value at the identity is $\gamma_G(x,y,z) = \kappa\bigl([x,y],z\bigr)$ for $x,y,z\in {\frak{g}}=T_eG$, where $\kappa$ is the Killing form of $\frak{g}$.  This form generates the deRham cohomology group $H^3_{dR}(G)\simeq \mathbb{R}$.  When $K=\mathrm{SO}(3)$ or $\mathrm{SU}(2)$, the Cartan $3$-form $\gamma_K$ is a canonical volume form, and, whenever $h:K\to G$ is a nonconstant homomorphism, one has $h^*(\gamma_G) = n_h\ \gamma_K$ for some integer $n_h>0$.
Now, consider the mapping $\mu:K\to G=\mathrm{SU}(n)$ defined by
$$
\mu(k) = (p_1k)(p_2k)\cdots(p_mk).
$$
By hypothesis, $\mu(k) = 1$ for all $k\in K$.  However, it is straightforward to compute that 
$$
\mu^*\bigl([\gamma_G]\bigr) = m\ n_h\ [\gamma_K]\not = 0,
$$ 
so it is not possible for $\mu$ to be constant.
