# AbstractsMathematics

by Glenn Steven Weeman

Institution: University of Akron Mathematics MS 2014 Mathematics; diophantine equation perfect group 2025066 http://rave.ohiolink.edu/etdc/view?acc_num=akron1396902470

## Abstract

Suppose that $G$ is a finite perfect group containing an involution $\tau$ whose centralizer subgroup in $G$ is a dihedral group of order $2^{d+1}$ where $d$ is an integer satisfying $d\geq 2$. Strayer proved in his Master's Thesis [1] that $|G|$ satisfies the equation $(a^2+(6b-8)a+b^2)|G|=2^{3d+2}a(a+1)$ where $\chi(1)=a$ and $\chi(\tau)=b$ for some ordinary irreducible character $\chi$ of $G$. From this it follows that $a$ is a positive integer and $b$ is an integer. Strayer also proved that the centralizer of $\tau$ is a Sylow 2-subgroup of $G$. From this it follows that the order of G is of the form $2^{d+1}c$ for some odd positive integer $c$. Hence the preceding equation is equivalent to $(a^2+(6b-8)a+b^2)c=2^{2d+1}a(a+1).$ We show that if there is a solution to the preceding equation of the form $(a,b,c,d)$ such that $b\not\in\{0,1,2,3\}$, then it is of the form $(a^*, 16k+3, c, d)$, $(a^{**}, 16k, c, d)$, $(16j+1, 2^{n+2}k+7\cdot 2^{n-1}+1, c, d)$, or $(16j-2, 2^{n+2}k+2^{n-1}+2, c, d)$ where $a^*$ is of one of the forms $2^{n+2}j+7\cdot 2^{n-1}-1$, $2^{m+1}j+2^{m-1}-1$, $2^{m+1}j+2^m-1$, or $2^{m+2}j+3\cdot 2^{m-1}-1$ and $a^{**}$ is of one of the forms $2^{n+2}j+2^{n-1}$, $2^{m+1}j+2^m$, $2^{m+1}j+3\cdot 2^{m-1}$, or $2^{m+2}j+5\cdot 2^{m-1}$ for a nonnegative integer $j$, an integer $k$, and even integers $m$ and $n$ satisfying $m\geq 14$ and $n\geq 4$. Furthermore, we show that $a\neq 1$, $a\neq 8$, $b\neq -7$, $b\neq 10$, and $d\geq 8$.