Subgroup $H$ of $\mathrm{GL}_n(\mathbf{R})$ where elements are of $H$ are matrices with only positive entries Suppose $H$ is a subgroup of $\mathrm{GL}_n(\mathbf{R})$ with the property that any element $h$ of $H$ is represented as a matrix with non-negative entries.
Two examples of $H$ I can think of include the group of diagonal matrices with positive entires and the group of permutation matrices. Let $K$ be the group generated by these two subgroups.
Because the formula for the inverse of a 2x2 matrix is available, as the inverse of any element of $H$ must also be in $H$, I can show that any such $H$ must be a subgroup of $K$ in this case.
My question is whether for $n \geq 3$ that $K$ contains any such $H$. 
 A: Think geometrically . . .

Let $H$ be such a group.

Any matrix $A\in H$ must map the set of standard unit basis vectors into the nonnegative orthant, but moreover, the image of the nonnegative orthant must be the nonnegative orthant, else $A^{-1}$ won't take the set of standard basis vectors into the nonnegative orthant.

It follows that $A$ maps each standard unit basis vector to a positive scalar multiple of some standard unit basis vector. Hence, up to a positive scale factor, $A$ permutes the standard unit basis vectors.

Thus, each $A\in H$ has exactly one positive entry in each row, and exactly one positive entry in each column, with all other entries equal to zero.

It follows that $H$ is a subgroup of the group $K$.
A: Proffering the following argument for an affirmative answer.
Let $G$ be such a group. Assume that $A=(a_{ij})\in G$ has two or more non-zero entries on some row, say, both $a_{ij}$ and $a_{ik}$ are positive. Let $B=(b_{ij})\in G$ be any matrix. Because $B$ is invertible there exists an $\ell$ such that $b_{\ell i}>0$. This implies that in the product $BA$ the entries $(\ell,j)$ and $(\ell,k)$ are both positive.
But this means that $BA$ cannot be the identity matrix contradicting the fact that $A^{-1}\in G$.
An analogous argument proves that no column of an element of $G$ can contain two or more non-zero entries. Therefore all the matricess of $G$ are monomial, and we are done.
