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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.

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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).

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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 $j = n$.

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