# The center of the group of $n\times n$ upper triangular matrices with a diagonal of ones

Let $$\mathbb{F}_{p}$$ be a finite field of order $$p$$ and $$H_{n}(\mathbb{F}_{p})$$ be the subgroup of $$GL_n(\mathbb{F}_{p})$$ of upper triangular matrices with a diagonal of ones. Note that the center $$Z(H_{3}(\mathbb{F}_{p}))$$ is well known and isomorphic to $$\mathbb{F}_{p}$$ (see center or dummit). Here, I'm looking for $$Z(H_{n}(\mathbb{F}_{p}))$$.

Any help would be appreciated so much. Thank you all.

The centre consists of upper-triangular matrices whose nonzero entries off the main diagonal are at the right upper corner. See Exercise 4. p 95 of (M. Suzuki, Group theory I, Springer Verlag, Berlin, 1982).

The center is isomorphic to a copy of $$\mathbb{F}_p$$ coming from the upper right corner.

Proof: for $$j > i$$ write $$E_{ij}$$ for the matrix which has a $$1$$ in spot $$(i, j)$$ and along the diagonal and is zero elsewhere.

Claim: if $$M$$ is a matrix with a nonzero entry somewhere other than the upper right corner, then there exist a choice of $$i, j$$ such that $$M$$ does not commute with $$E_{ij}$$.

Now left-multiplication by $$E_{ij}$$ replaces $$M$$ with the matrix $$M^L$$ whose $$i$$th row is the sum of rows $$i$$ and $$j$$ of $$M$$, and right-multiplication by $$E_{ij}$$ replaces $$M$$ with the matrix $$M^R$$ whose $$j$$th column is the sum of columns $$i$$ and $$j$$ of $$M$$.

For these two matrices to be the same, there must be no non-zero entry of column $$i$$ outside row $$i$$ (otherwise $$M^R$$ will have column $$j$$ different from $$M^L$$, since the only entry of column $$j$$ of $$M^L$$ which is different from that of $$M$$ is the one in row $$i$$).

Varying $$i$$, we see that all entries of $$M$$ must be zero except the diagonal ones in all columns except column $$n$$ (since then we can't take $$j > i$$). The same argument on the rows eliminates all rows except $$j = n$$.